MHT CET 2025 Chemistry Question Paper with Answer and Solution

(A) The energy of one photon is given by $E = \frac{hc}{\lambda}$.
Given: $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m/s$,$\lambda = 700 \ nm = 700 \times 10^{-9} \ m$.
Energy of one photon $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} \ J = 2.84 \times 10^{-19} \ J$.
To find the energy per mole,multiply by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
$E_{mole} = E \times N_A = 2.84 \times 10^{-19} \times 6.022 \times 10^{23} \ J/mol \approx 1.71 \times 10^5 \ J/mol$.

(B) The relationship between frequency $(\nu)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Given:
$c = 3 \times 10^8 \ ms^{-1}$
$\lambda = 600 \ nm = 600 \times 10^{-9} \ m = 6 \times 10^{-7} \ m$.
Substituting the values into the formula:
$\nu = \frac{3 \times 10^8 \ ms^{-1}}{6 \times 10^{-7} \ m} = 0.5 \times 10^{15} \ Hz = 5.0 \times 10^{14} \ Hz$.
Therefore,the correct option is $B$.

(A) According to Bohr's postulate,the angular momentum $(L)$ of an electron in an orbit is given by the formula: $L = \frac{nh}{2\pi}$.
For the first orbit,$n = 1$.
Substituting the values: $L = \frac{1 \times 6.626 \times 10^{-34}}{2 \times 3.14159}$.
$L = \frac{6.626 \times 10^{-34}}{6.28318} \approx 1.0545 \times 10^{-34} \ J \ s$.
Thus,the correct option is $A$.

(D) . The statement "It is impossible to determine simultaneously the exact position and exact momentum of an electron" is the definition of the $Heisenberg \ uncertainty \ principle$.
According to this principle,the product of the uncertainty in position $(\Delta x)$ and the uncertainty in momentum $(\Delta p)$ is greater than or equal to $\frac{h}{4\pi}$,expressed as $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.

(C) According to the de Broglie equation,$\lambda = \frac{h}{mv}$.
Rearranging for velocity,we get $v = \frac{h}{m \lambda}$.
Given values: $h = 6.63 \times 10^{-34} \ Js$,$m = 9.1 \times 10^{-31} \ kg$,and $\lambda = 58 \ nm = 58 \times 10^{-9} \ m$.
Substituting these values into the formula:
$v = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (58 \times 10^{-9})}$
$v = \frac{6.63 \times 10^{-34}}{527.8 \times 10^{-40}}$
$v = \frac{6.63}{527.8} \times 10^6 \ ms^{-1}$
$v \approx 0.01256 \times 10^6 \ ms^{-1} = 1.256 \times 10^4 \ ms^{-1}$.
Rounding to the nearest option,the velocity is $1.26 \times 10^4 \ ms^{-1}$.

(B) The number of radial nodes in an orbital is given by the formula: $\text{Radial nodes} = n - l - 1$.
For the $2s$ orbital,the principal quantum number $n = 2$ and the azimuthal quantum number $l = 0$.
Substituting these values into the formula: $\text{Radial nodes} = 2 - 0 - 1 = 1$.
Therefore,the $2s$ orbital has $1$ radial node.

(D) The $d$-orbitals are five in number: $d_{xy}$,$d_{yz}$,$d_{xz}$,$d_{x^2-y^2}$,and $d_{z^2}$.
Among these,the first four ($d_{xy}$,$d_{yz}$,$d_{xz}$,and $d_{x^2-y^2}$) have a double-dumbbell shape.
The $d_{z^2}$ orbital has a unique shape consisting of a dumbbell along the $z$-axis with a doughnut-shaped ring of electron density in the $xy$-plane.
Therefore,$d_{z^2}$ has a different shape compared to the others.

(A) The statement "No two electrons in an atom can have the identical set of four quantum numbers" is the definition of $Pauli's \ exclusion \ principle$.
According to this principle,each orbital can hold a maximum of two electrons,and these two electrons must have opposite spins.

(C) To find the number of unpaired electrons,we write the electronic configuration for each element:
$1$. Fluorine ($F$,$Z=9$): $1s^2 2s^2 2p^5$. The $2p$ subshell has $5$ electrons,resulting in $1$ unpaired electron.
$2$. Sodium ($Na$,$Z=11$): $1s^2 2s^2 2p^6 3s^1$. The $3s$ subshell has $1$ unpaired electron.
$3$. Nitrogen ($N$,$Z=7$): $1s^2 2s^2 2p^3$. According to Hund's rule,the $2p$ subshell has $3$ electrons in separate orbitals,resulting in $3$ unpaired electrons.
$4$. Oxygen ($O$,$Z=8$): $1s^2 2s^2 2p^4$. The $2p$ subshell has $4$ electrons,resulting in $2$ unpaired electrons.
Comparing the values,Nitrogen has the maximum number of unpaired electrons $(3)$.

(B) The statement describes $Hund's \ rule \ of \ maximum \ multiplicity$.
According to this rule,for a given electron configuration,the term with maximum multiplicity has the lowest energy.
In simpler terms,pairing of electrons in the orbitals of a given subshell $(p, d, f)$ does not take place until each orbital is singly occupied.

(A) The given reaction is the water-gas shift reaction.
In this industrial process,$CO$ reacts with steam at $623 \ K$ in the presence of an iron chromate $(Fe_2O_3 \cdot Cr_2O_3)$ catalyst to produce $CO_2$ and $H_2$.

(A) An intensive property is a physical property of a system that does not depend on the amount of matter present or the size of the system.
$1$. Surface tension is an intensive property because it depends only on the nature of the substance and temperature,not on the quantity of the liquid.
$2$. Volume,internal energy,and the number of moles are extensive properties because they depend on the amount of matter present in the system.

(C) $1$. An intensive property is a property that does not depend on the amount of matter present in the system (e.g.,$Temperature$,$Pressure$,$Density$).
$2$. $A$ state function is a property whose value depends only on the state of the system and not on the path taken to reach that state (e.g.,$Internal \ energy$,$Enthalpy$,$Entropy$,$Temperature$).
$3$. $Internal \ energy$,$Volume$,and $Entropy$ are extensive properties because they depend on the amount of matter.
$4$. $Temperature$ is independent of the amount of matter (intensive) and its value depends only on the current state of the system (state function).
$5$. Therefore,$Temperature$ is the correct answer.

(D) An intensive property is a physical property of a system that does not depend on the amount of matter present or the size of the system.
$Volume$,$Enthalpy$,and $Entropy$ are extensive properties because they depend on the quantity of matter.
$Molar volume$ is defined as the volume per mole of a substance $(V_m = V/n)$,which is independent of the total amount of the substance.
Therefore,$Molar volume$ is an intensive property.

(A) Work done in a chemical reaction at constant pressure is given by $W = -P \Delta V = -\Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For zero work,$\Delta n_g$ must be equal to $0$.
In option $A$: $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
In option $B$: $\Delta n_g = 2 - (3 + 1) = -2$.
In option $C$: $\Delta n_g = 2 - (1 + 2.5) = -1.5$.
In option $D$: $\Delta n_g = 4 - (2 + 7) = -5$.
Since $\Delta n_g = 0$ for reaction $A$,it performs zero work.

(B) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done on the system,so $w = +20 \ kJ$.
The system releases heat,so $q = -10 \ kJ$.
Substituting these values into the equation: $\Delta U = -10 \ kJ + 20 \ kJ = +10 \ kJ$.
Therefore,the change in internal energy is $10 \ kJ$.

(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ext} \Delta V$.
Given: $W = -600 \ J$,$V_1 = 15 \ dm^{3}$,$V_2 = 20 \ dm^{3}$.
Change in volume $\Delta V = V_2 - V_1 = 20 \ dm^{3} - 15 \ dm^{3} = 5 \ dm^{3}$.
Since $1 \ dm^{3} = 1 \ L$ and $1 \ L \cdot bar = 100 \ J$,we convert the volume change to units compatible with Joules.
$5 \ dm^{3} = 5 \ L$.
Using $W = -P_{ext} \Delta V$,we have $-600 \ J = -P_{ext} \times (5 \ L)$.
$P_{ext} = \frac{600 \ J}{5 \ L} = 120 \ J/L$.
Since $100 \ J = 1 \ L \cdot bar$,then $120 \ J = 1.2 \ L \cdot bar$.
Therefore,$P_{ext} = \frac{120 \ J}{5 \ L} = 24 \ J/L = 0.24 \ bar$.
Wait,re-evaluating the conversion: $1 \ dm^{3} \cdot bar = 100 \ J$.
$P_{ext} (bar) = \frac{-W (J)}{100 \times \Delta V (dm^{3})} = \frac{600}{100 \times 5} = \frac{600}{500} = 1.2 \ bar$.
The correct answer is $1.2 \ bar$.

(D) The work done in a thermodynamic process is given by the formula $W = -P_{ext} \Delta V$.
In the case of free expansion,a gas expands against a vacuum,meaning the external pressure $P_{ext} = 0$.
Therefore,the work done $W = -0 \times \Delta V = 0$.
Thus,free expansion involves zero work done.

(A) For an isothermal reversible compression of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \log\left(\frac{V_f}{V_i}\right)$.
Given: $n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_i = 12 \ dm^3$,$V_f = 6 \ dm^3$.
Substituting the values: $w = -2.303 \times 1 \times 8.314 \times 300 \times \log\left(\frac{6}{12}\right)$.
$w = -2.303 \times 8.314 \times 300 \times \log(0.5)$.
Since $\log(0.5) = -0.3010$,we have: $w = -2.303 \times 8.314 \times 300 \times (-0.3010)$.
$w \approx 1728.8 \ J = 1.729 \ kJ$.

(A) The work done in a chemical reaction is given by the formula $W = -\Delta n_g RT$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g = n_{p(g)} - n_{r(g)}$.
For the reaction $4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$,the number of moles of gaseous products is $4$ and the number of moles of gaseous reactants is $4 + 2 = 6$.
So,$\Delta n_g = 4 - 6 = -2 \ mol$.
The temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
The gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Now,substitute the values into the formula: $W = -(-2 \ mol) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K)$.
$W = 2 \times 8.314 \times 300 = 4988.4 \ J$.

(C) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,$q$ is the heat absorbed by the system and $w$ is the work done on the system.
Given:
Heat absorbed by the system,$q = +10 \ kJ$ (positive because heat is absorbed).
Work done by the system,$w = -25 \ kJ$ (negative because work is done by the system).
Substituting these values into the equation:
$\Delta U = 10 \ kJ + (-25 \ kJ) = -15 \ kJ$.
Therefore,the change in internal energy is $-15 \ kJ$.

(A) The work done during the expansion of a gas against a constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 1 \ atm$,$V_1 = 15.5 \ dm^3$,and $V_2 = 20 \ dm^3$.
Change in volume,$\Delta V = V_2 - V_1 = 20 \ dm^3 - 15.5 \ dm^3 = 4.5 \ dm^3$.
Since $1 \ dm^3 \cdot atm = 101.325 \ J$,the work done is $W = -1 \ atm \times 4.5 \ dm^3 = -4.5 \ dm^3 \cdot atm$.
Converting to joules: $W = -4.5 \times 101.325 \ J \approx -456 \ J$.

(C) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done on the system,so $w = +20 \ kJ$.
The system releases heat,so $q = -10 \ kJ$.
Substituting these values into the equation: $\Delta U = -10 \ kJ + 20 \ kJ = +10 \ kJ$.
Therefore,the change in internal energy is $10 \ kJ$.

(A) The work done by a system during a chemical reaction is given by the formula $W = -P \Delta V = -\Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species.
If $\Delta n_g > 0$,then $W < 0$ (work is done by the system).
If $\Delta n_g < 0$,then $W > 0$ (work is done on the system).
For option $A$: $\Delta n_g = 1 - 0 = 1$. Since $\Delta n_g > 0$,work is done by the system $(W < 0)$.
For option $B$: $\Delta n_g = 0 - 2 = -2$. Since $\Delta n_g < 0$,work is done on the system $(W > 0)$.
For option $C$: $\Delta n_g = 2 - 2 = 0$. Since $\Delta n_g = 0$,$W = 0$.
For option $D$: $\Delta n_g = 2 - 4 = -2$. Since $\Delta n_g < 0$,work is done on the system $(W > 0)$.
Therefore,the reaction exhibiting negative work done by the system is $2 H_2 O_{2(\ell)} \rightarrow 2 H_2 O_{(\ell)} + O_{2_{(g)}}$.

(C) For an isothermal reversible expansion,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log\left(\frac{P_1}{P_2}\right)$.
Given values are: $n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 10 \ bar$,and $P_2 = 1 \ bar$.
Substituting these values into the equation:
$w = -2.303 \times 1 \times 8.314 \times 300 \times \log\left(\frac{10}{1}\right)$.
Since $\log(10) = 1$,we have:
$w = -2.303 \times 8.314 \times 300 \times 1$.
$w = -5744.14 \ J$.
Converting to $kJ$,we get $w = -5.744 \ kJ \approx -5.74 \ kJ$.

(A) The work done in an irreversible isothermal expansion against constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Given:
$P_{ext} = 1 \ atm$
$V_1 = 2 \ dm^3$
$V_2 = 2.8 \ dm^3$
$\Delta V = V_2 - V_1 = 2.8 \ dm^3 - 2 \ dm^3 = 0.8 \ dm^3$.
Since $1 \ dm^3 \cdot atm = 101.325 \ J$,
$W = -1 \ atm \times 0.8 \ dm^3 = -0.8 \ dm^3 \cdot atm$.
$W = -0.8 \times 101.325 \ J = -81.06 \ J$.
Rounding to the nearest option,the work done is $-81.04 \ J$.

(B) The work done in an irreversible process against constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 4 \ bar = 4 \times 10^5 \ Pa$.
Initial volume $V_1 = 25 \ dm^3 = 25 \times 10^{-3} \ m^3$.
Final volume $V_2 = 13 \ dm^3 = 13 \times 10^{-3} \ m^3$.
Change in volume $\Delta V = V_2 - V_1 = (13 - 25) \times 10^{-3} \ m^3 = -12 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(4 \times 10^5 \ Pa) \times (-12 \times 10^{-3} \ m^3) = 4800 \ J$.
Since the gas is compressed,work is done on the system,so the value is positive.

$(A)$ The formula for work done in an irreversible process against constant external pressure is given by $W = -P_{ext} \times \Delta V$.
Here, $P_{ext} = 3 \ bar = 3 \times 10^5 \ Pa$.
$\Delta V = V_f - V_i = 13 \ dm^3 - 24 \ dm^3 = -11 \ dm^3 = -11 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(3 \times 10^5 \ Pa) \times (-11 \times 10^{-3} \ m^3)$.
$W = 3300 \ J$.
Since the gas is compressed, work is done on the system, hence the value is positive.

(B) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,the system absorbs heat,so $q = +50 \ J$.
The work is done by the system,so $w = -18 \ J$.
Substituting these values into the equation: $\Delta U = 50 \ J + (-18 \ J) = 32 \ J$.
Therefore,the change in internal energy is $32 \ J$.

(A) For an isothermal reversible compression,the work done $w$ is given by the formula: $w = -nRT \ln\left(\frac{P_2}{P_1}\right)$.
Given values: $n = 1 \ mol$,$T = 300 \ K$,$P_1 = x \ bar$,$P_2 = 2x \ bar$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the formula:
$w = -(1 \ mol) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K) \times \ln\left(\frac{2x}{x}\right)$.
$w = -8.314 \times 300 \times \ln(2)$.
Using $\ln(2) \approx 0.693$:
$w = -2494.2 \times 0.693 \approx -1728.5 \ J$.
Converting to $kJ$: $w \approx -1.729 \ kJ$.
The magnitude of work done on the system is $1.729 \ kJ$.

(B) For an isothermal reversible expansion of an ideal gas,the work done $W$ is given by the formula: $W = -2.303 \ nRT \ \log(\frac{P_1}{P_2})$
Given:
$n = 3 \ mol$
$T = 27^{\circ} C = 27 + 273 = 300 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$P_1 = 10 \ atm$
$P_2 = 1 \ atm$
Substituting the values:
$W = -2.303 \times 3 \times 8.314 \times 300 \times \log(\frac{10}{1})$
$W = -2.303 \times 3 \times 8.314 \times 300 \times 1$
$W = -17234.6 \ J$
Converting to $kJ$:
$W = -17.2346 \ kJ \approx -17.23 \ kJ$

(B) The reaction is $4 HCl_{(g)} + O_{2_{(g)}} \rightarrow 2 Cl_{2_{(g)}} + 2 H_{2}O_{(g)}$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g = (n_{products} - n_{reactants}) = (2 + 2) - (4 + 1) = 4 - 5 = -1$.
This $\Delta n_g$ is for $4 \ moles$ of $HCl$.
For $1 \ mole$ of $HCl$,$\Delta n_g = -1 / 4 = -0.25$.
The work done is given by $W = -\Delta n_g RT$.
Given $T = 27^{\circ} C = 300 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$W = -(-0.25) \times 8.314 \times 300$.
$W = 0.25 \times 2494.2 = 623.55 \ J \approx 623.6 \ J$.

(C) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g \ mol^{-1}$.
Number of moles of benzene $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{13 \ g}{78 \ g \ mol^{-1}} = \frac{1}{6} \ mol$.
The enthalpy of vaporisation $(\Delta_{vap}H)$ is given by the heat supplied per mole: $\Delta_{vap}H = \frac{q}{n}$.
$\Delta_{vap}H = \frac{5.1 \ kJ}{1/6 \ mol} = 5.1 \times 6 \ kJ \ mol^{-1} = 30.6 \ kJ \ mol^{-1}$.

(B) The molar mass of ethanol $(C_2H_5OH)$ is calculated as: $(2 \times 12.01) + (6 \times 1.008) + (1 \times 16.00) = 46.07 \ g \ mol^{-1}$,which is approximately $46 \ g \ mol^{-1}$.
Number of moles of ethanol $(n) = \frac{\text{mass}}{\text{molar mass}} = \frac{11.5 \ g}{46 \ g \ mol^{-1}} = 0.25 \ mol$.
The enthalpy of vaporisation $(\Delta_{vap}H)$ is the heat required to vaporise $1 \ mol$ of a substance.
$\Delta_{vap}H = \frac{\text{Heat supplied}}{\text{Number of moles}} = \frac{11.8 \ kJ}{0.25 \ mol} = 47.2 \ kJ \ mol^{-1}$.
Therefore,the correct option is $B$.

(A) The standard enthalpy change of the reaction is calculated using the formula:
$\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$
For the reaction: $C_2H_{4(g)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 2H_2O_{(\ell)}$
$\Delta_{r}H^{\circ} = [2 \times \Delta_{f}H^{\circ}(CO_2) + 2 \times \Delta_{f}H^{\circ}(H_2O)] - [1 \times \Delta_{f}H^{\circ}(C_2H_4) + 3 \times \Delta_{f}H^{\circ}(O_2)]$
Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state):
$\Delta_{r}H^{\circ} = [2(-393.5) + 2(-285.8)] - [52 + 0]$
$\Delta_{r}H^{\circ} = [-787 - 571.6] - 52$
$\Delta_{r}H^{\circ} = -1358.6 - 52 = -1410.6 \ kJ \ mol^{-1}$
The closest value is $-1411.1 \ kJ \ mol^{-1}$.

(C) The standard enthalpy of formation $(\Delta_{f}H^0)$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the enthalpy change is $\Delta_{r}H^0 = -92.0 \ kJ$ for the production of $2 \ mol$ of $NH_3$.
To find the enthalpy of formation per mole of $NH_3$,we divide the reaction enthalpy by the stoichiometric coefficient of $NH_3$:
$\Delta_{f}H^0(NH_3) = \frac{\Delta_{r}H^0}{2} = \frac{-92.0 \ kJ}{2} = -46.0 \ kJ \ mol^{-1}$.

(C) The standard enthalpy change of reaction $\Delta_rH^{\circ}$ is calculated using the formula:
$\Delta_rH^{\circ} = \sum \Delta_fH^{\circ}(\text{products}) - \sum \Delta_fH^{\circ}(\text{reactants})$
For the reaction $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(\ell)}$:
$\Delta_rH^{\circ} = [2 \times \Delta_fH^{\circ}(CO_2) + 1 \times \Delta_fH^{\circ}(H_2O)] - [1 \times \Delta_fH^{\circ}(C_2H_2) + \frac{5}{2} \times \Delta_fH^{\circ}(O_2)]$
Since $\Delta_fH^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element):
$\Delta_rH^{\circ} = [2(-393) + (-286)] - [227 + 0]$
$\Delta_rH^{\circ} = [-786 - 286] - 227$
$\Delta_rH^{\circ} = -1072 - 227 = -1299 \ kJ \ mol^{-1}$

(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})$
For the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$,the expression is:
$\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_2) + 2 \times \Delta_{f} H^{\circ}(H_2O)] - [\Delta_{f} H^{\circ}(CH_4) + 2 \times \Delta_{f} H^{\circ}(O_2)]$
Since $O_2$ is an element in its standard state,$\Delta_{f} H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$.
Substituting the given values:
$\Delta_{r} H^{\circ} = [-394 + 2 \times (-286)] - [-75 + 0]$
$\Delta_{r} H^{\circ} = [-394 - 572] - [-75]$
$\Delta_{r} H^{\circ} = -966 + 75 = -891 \ kJ \ mol^{-1}$

(B) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Number of moles of water $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{1.8 \ g}{18 \ g \ mol^{-1}} = 0.1 \ mol$.
Heat supplied $(q)$ = $4 \ kJ$.
The molar heat of vaporization $(\Delta H_{vap})$ is defined as the heat required to vaporize $1 \ mole$ of a substance.
$\Delta H_{vap} = \frac{q}{n} = \frac{4 \ kJ}{0.1 \ mol} = 40 \ kJ \ mol^{-1}$.
Therefore,the correct option is $B$.

(A) The enthalpy of solution $(\Delta_{sol} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Delta_{hyd} H)$.
$\Delta_{sol} H = \Delta_{L} H + \Delta_{hyd} H$
Given:
$\Delta_{L} H = 700 \ kJ \ mol^{-1}$
$\Delta_{hyd} H = -680 \ kJ \ mol^{-1}$
Substituting the values:
$\Delta_{sol} H = 700 \ kJ \ mol^{-1} + (-680 \ kJ \ mol^{-1})$
$\Delta_{sol} H = 20 \ kJ \ mol^{-1}$
Therefore,the correct option is $A$.

(C) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$.
For the reaction $C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$,the expression is:
$\Delta_{r}H^{\circ} = [2 \times \Delta_{f}H^{\circ}(CO_2) + 3 \times \Delta_{f}H^{\circ}(H_2O)] - [\Delta_{f}H^{\circ}(C_2H_5OH) + 3 \times \Delta_{f}H^{\circ}(O_2)]$.
Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we have:
$\Delta_{r}H^{\circ} = [2(-390) + 3(-285)] - [-280 + 3(0)]$.
$\Delta_{r}H^{\circ} = [-780 - 855] - [-280]$.
$\Delta_{r}H^{\circ} = -1635 + 280 = -1355 \ kJ \ mol^{-1}$.

(A) An exothermic reaction is one that releases heat to the surroundings,characterized by a negative enthalpy change $(\Delta H < 0)$.
$A$. The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is a neutralization reaction. Neutralization reactions are always exothermic because they involve the formation of water from $H^+$ and $OH^-$ ions,releasing energy.
$B$. The melting of ice is an endothermic process as it requires the absorption of heat to break intermolecular hydrogen bonds.
$C$. The dissolution of $NaCl$ in water is an endothermic process (or slightly endothermic) as it requires energy to overcome the lattice enthalpy of the crystal.
$D$. The formation of nitrogen dioxide from nitrogen and oxygen is an endothermic reaction,requiring energy input.
Therefore,the correct option is $A$.

(A) The synthesis of ammonia gas is represented by the equation:
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
To obtain this equation,we add the two given reactions:
Reaction $i$: $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
Reaction $ii$: $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$
Adding reaction $i$ and reaction $ii$:
$(2 H_{2(g)} + N_{2(g)}) + (N_{2}H_{4(g)} + H_{2(g)}) \longrightarrow N_{2}H_{4(g)} + 2 NH_{3(g)}$
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
The total enthalpy change is $\Delta_{r}H^{0} = \Delta_{r}H_{1}^{0} + \Delta_{r}H_{2}^{0}$
$\Delta_{r}H^{0} = 95.4 \ kJ + (-187.6 \ kJ) = -92.2 \ kJ$
This is the enthalpy change for the production of $2 \ moles$ of $NH_{3}$.
Therefore,the correct option is $A$.

(B) An endothermic process is one that absorbs heat from the surroundings.
Phase transitions that involve moving from a more ordered state to a less ordered state (e.g.,solid to liquid or liquid to gas) require energy input to overcome intermolecular forces.
In the transformation $H_2O_{(s)} \rightarrow H_2O_{(\ell)}$,ice melts into liquid water,which is an endothermic process because heat is absorbed to break the crystal lattice of ice.
The other options ($H_2O_{(\ell)} \rightarrow H_2O_{(s)}$,$H_2O_{(g)} \rightarrow H_2O_{(\ell)}$,and $H_2O_{(g)} \rightarrow H_2O_{(s)}$) represent freezing or condensation,which are exothermic processes as they release heat.

(A) The standard enthalpy change of reaction $\Delta_{r} H^{\circ}$ is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})$.
For the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$,the expression is: $\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_{2}) + 2 \times \Delta_{f} H^{\circ}(H_{2}O)] - [\Delta_{f} H^{\circ}(CH_{4}) + 2 \times \Delta_{f} H^{\circ}(O_{2})]$.
Since $\Delta_{f} H^{\circ}(O_{2}) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we substitute the given values:
$\Delta_{r} H^{\circ} = [-390 + 2 \times (-286)] - [-75 + 2 \times 0]$
$\Delta_{r} H^{\circ} = [-390 - 572] - [-75]$
$\Delta_{r} H^{\circ} = -962 + 75 = -887 \ kJ \ mol^{-1}$.

(C) The reaction $C_{(graphite)} \longrightarrow C_{(g)}$ represents the conversion of a solid substance (graphite) directly into its gaseous state.
This process of phase transition from solid to gas is known as sublimation.
Therefore,the correct process is sublimation.

(A) The total entropy change is given by the formula: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
Given $\Delta S_{sys} = \Delta S^{\circ} = -36 \ J \ K^{-1}$.
The entropy change of the surroundings is calculated as: $\Delta S_{surr} = -\frac{\Delta H^{\circ}}{T}$.
Convert $\Delta H^{\circ}$ to $J$: $\Delta H^{\circ} = -208.6 \ kJ = -208600 \ J$.
$\Delta S_{surr} = -\frac{-208600 \ J}{298 \ K} = 700 \ J \ K^{-1}$.
Therefore,$\Delta S_{total} = -36 \ J \ K^{-1} + 700 \ J \ K^{-1} = 664 \ J \ K^{-1}$.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. A decrease in entropy $(\Delta S < 0)$ occurs when the number of moles of gaseous products is less than the number of moles of gaseous reactants, or when a gas is converted into a liquid or solid.
$A$: $2H_2O_{2(l)} \longrightarrow 2H_2O_{(l)} + O_{2(g)}$. Here, $0$ moles of gas become $1$ mole of gas. $\Delta S > 0$.
$B$: $H_{2(g)} \longrightarrow 2H_{(g)}$. Here, $1$ mole of gas becomes $2$ moles of gas. $\Delta S > 0$.
$C$: $CaCO_{3(s)} \longrightarrow CaO_{(s)} + CO_{2(g)}$. Here, $0$ moles of gas become $1$ mole of gas. $\Delta S > 0$.
$D$: $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$. Here, $3$ moles of gas become $0$ moles of gas (liquid). $\Delta S < 0$.
Therefore, reaction $D$ exhibits a decrease in entropy.

(B) The entropy change of the surrounding $(\Delta S_{surr})$ is given by the formula: $\Delta S_{surr} = \frac{-q_{sys}}{T}$.
Given that the reaction releases heat,$q_{sys} = -525 \ kJ = -525000 \ J$.
Therefore,the heat absorbed by the surrounding is $q_{surr} = +525000 \ J$.
The temperature $T = 300 \ K$.
Substituting the values: $\Delta S_{surr} = \frac{525000 \ J}{300 \ K} = 1750 \ J \ K^{-1}$.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system.
For a given substance,the entropy follows the order: $S_{(gas)} > S_{(liquid)} > S_{(solid)}$.
$A$ decrease in entropy occurs when the system moves from a state of higher disorder to a state of lower disorder.
In the process $H_2O_{(g)} \longrightarrow H_2O_{(\ell)}$,the substance changes from a gaseous state (high entropy) to a liquid state (lower entropy).
Therefore,this transformation exhibits a decrease in entropy.

(D) The atomic number of lutetium $(Lu)$ is $71$.
The electronic configuration of neutral $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$.
In its $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$),resulting in the configuration $[Xe] 4f^{14}$.
Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons in the $f$ orbitals.
Therefore,the number of unpaired electrons is $0$.

(A) In the lanthanoid series,the basic strength of hydroxides decreases as the atomic number increases from $La$ to $Lu$. This is due to the lanthanoid contraction,which leads to a decrease in ionic radius and an increase in covalent character of the $M-OH$ bond. Since $La^{3+}$ has the largest ionic radius among the given options,the $La-OH$ bond is the most ionic,making $La(OH)_3$ the strongest base.

(A) In the actinoid series,as the atomic number increases,the ionic radius decreases due to the actinoid contraction,which is analogous to lanthanoid contraction.
This occurs because the $5f$ electrons provide poor shielding for the increasing nuclear charge.
Therefore,the actinoid with the lowest atomic number will have the largest ionic radius in its $+3$ state.
Comparing the atomic numbers: $U$ $(Z=92)$,$Bk$ $(Z=97)$,$Es$ $(Z=99)$,and $Md$ $(Z=101)$.
Since $U$ has the lowest atomic number among the given options,it has the largest size in its $+3$ state.

(C) The electrolysis of aqueous $NaCl$ at the anode involves the oxidation of chloride ions: $2Cl^- (aq) \rightarrow Cl_2 (g) + 2e^-$.
From the stoichiometry,$1 \ mol$ of $Cl_2$ requires $2 \ mol$ of electrons.
Therefore,$0.5 \ mol$ of $Cl_2$ requires $0.5 \times 2 = 1 \ mol$ of electrons.
Using Faraday's law,the total charge $Q = n \times F$,where $n = 1 \ mol$ and $F = 96500 \ C \ mol^{-1}$.
So,$Q = 96500 \ C$.
Given current $I = 100 \ A$,we use the formula $Q = I \times t$.
$t = Q / I = 96500 \ C / 100 \ A = 965 \ seconds$.

(A) The Nernst equation for the cell reaction is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$,where $n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.
Given that $E_{cell} = E^{\circ}_{cell} - 0.0592 \ V$,we have:
$E^{\circ}_{cell} - \frac{0.0592}{2} \log Q = E^{\circ}_{cell} - 0.0592$
$\frac{1}{2} \log Q = 1 \implies \log Q = 2 \implies Q = 10^2 = 100$.
Checking the options:
For option $A$: $Q = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.
Thus,option $A$ is correct.

(C) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.
Initially,$Q_1 = \frac{1}{(1)^2} = 1$,so $E_1 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(1) = E^{\circ}_{cell}$.
After the concentration change,$Q_2 = \frac{0.1}{(1)^2} = 0.1$.
$E_2 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(0.1) = E^{\circ}_{cell} - 0.0296 \times (-1) = E^{\circ}_{cell} + 0.0296 \ V$.
The change in $EMF$ is $E_2 - E_1 = +0.0296 \ V$.
Thus,the $EMF$ increases by $0.0296 \ V$.

(B) The reduction reaction for $H^{+}$ ions to form $H_2$ gas is given by:
$2H^{+} + 2e^{-} \rightarrow H_2$
According to the stoichiometry of the reaction,$2 \ mol$ of electrons are required to produce $1 \ mol$ of $H_2$ gas.
Since $1 \ mol$ of electrons carries a charge of $1 \ Faraday$ $(F)$,the total charge required for $2 \ mol$ of electrons is $2 \ F$.

(D) In a Galvanic cell,the electrode with the higher reduction potential acts as the cathode (positive electrode),and the electrode with the lower reduction potential acts as the anode (negative electrode).
Given $E^{\circ}(Zn^{+2}/Zn) = -0.76 \ V$ and $E^{\circ}(H^{+}/H_2) = 0.00 \ V$.
Since $0.00 \ V > -0.76 \ V$,the standard hydrogen electrode acts as the cathode (positive electrode).
Reduction occurs at the cathode: $2H^{+}_{(aq)} + 2e^{-} \longrightarrow H_{2(g)}$.

(B) The relationship between the standard Gibbs free energy change $(\Delta G^{\circ})$ and the standard cell potential $(E_{\text{cell}}^{\circ})$ is given by the equation: $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$.
Rearranging this for $E_{\text{cell}}^{\circ}$,we get: $E_{\text{cell}}^{\circ} = \frac{-\Delta G^{\circ}}{nF}$.
In a dry cell (Leclanché cell),the overall reaction involves the reduction of $MnO_2$ and the oxidation of $Zn$. The reaction is: $Zn(s) + 2MnO_2(s) + 2NH_4^+(aq) \rightarrow Zn^{2+}(aq) + Mn_2O_3(s) + 2NH_3(g) + H_2O(l)$.
Here,the number of electrons transferred $(n)$ is $2$.
Substituting $n = 2$ into the formula,we get $E_{\text{cell}}^{\circ} = \frac{-\Delta G^{\circ}}{2F}$.

(D) The standard electrode potential $(E^\circ)$ is an intensive property,meaning it does not depend on the amount of substance present or the stoichiometric coefficients of the reaction.
For the reduction reaction: $Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$,$E^\circ = -0.76 \ V$.
The given reaction is the reverse of the reduction reaction multiplied by a factor of $2$: $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$.
Reversing the reaction changes the sign of the potential: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$,$E^\circ = -(-0.76 \ V) = +0.76 \ V$.
Since the potential is an intensive property,multiplying the reaction by a coefficient does not change the value of $E^\circ$.
Therefore,the standard potential for the reaction $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$ remains $+0.76 \ V$.

(D) The Nernst equation for the given cell reaction is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.
Here,the number of electrons transferred $n = 2$.
Given that $[Cd^{2+}] = 10 [Cu^{2+}]$,so the ratio $\frac{[Cd^{2+}]}{[Cu^{2+}]} = 10$.
Substituting these values into the equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log(10)$.
Since $\log(10) = 1$,we get $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \times 1$.
$E_{cell} = E^{\circ}_{cell} - 0.02955 \ V \approx E^{\circ}_{cell} - 0.0296 \ V$.
Therefore,the $emf$ of the cell is less than $E^{\circ}_{cell}$ by $0.0296 \ V$.

(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$.
Given: $\Delta G^{\circ} = -386 \ kJ = -386000 \ J$,$n = 2$,and $F \approx 96500 \ C \ mol^{-1}$.
Substituting the values: $-386000 \ J = -(2) \times (96500 \ C \ mol^{-1}) \times E_{\text{cell}}^{\circ}$.
$E_{\text{cell}}^{\circ} = \frac{386000}{2 \times 96500} = \frac{386000}{193000} = 2 \ V$.
Therefore,the correct option is $C$.

(C) The cell reaction is: $Zn_{(s)} + 2Ag_{(1 \ M)}^{+1} \rightarrow Zn_{(1 \ M)}^{+2} + 2Ag_{(s)}$.
Here, the number of electrons transferred $(n)$ is $2$.
The standard $emf$ of the cell $(E_{cell}^o)$ is $1.55 \ V$.
The electrical work done under standard conditions is equal to the change in Gibbs free energy $(\Delta G^o)$.
$\Delta G^o = -nFE_{cell}^o$.
Substituting the values: $n = 2$, $F = 96500 \ C \ mol^{-1}$, $E_{cell}^o = 1.55 \ V$.
$\Delta G^o = -2 \times 96500 \times 1.55 = -299150 \ J = -299.150 \ kJ$.
Therefore, the electrical work done is $-299.150 \ kJ$.

(A) The given reaction is the oxidation of $Mg$ to $Mg^{2+}$.
The standard electrode potential for reduction is $E^{\circ} (Mg^{2+} \mid Mg) = -2.37 \ V$.
Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$.
Using the Nernst equation for the reaction $Mg_{(s)} \longrightarrow Mg^{2+} (0.1 \ M) + 2 \ e^{-}$:
$E = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{2+}]$
$E = 2.37 - \frac{0.0591}{2} \log (0.1)$
$E = 2.37 - 0.02955 \times (-1)$
$E = 2.37 + 0.02955 = 2.39955 \ V \approx +2.3996 \ V$.

(C) $1$. Standard cell potential $(E^{\circ}_{cell})$ is an intensive property because it does not depend on the amount of matter present in the system.
$2$. Electrode potential depends on the concentration of the electrolyte solution as described by the Nernst equation $(E = E^{\circ} - \frac{RT}{nF} \ln Q)$.
$3$. Standard free energy change $(\Delta G^{\circ} = -nFE^{\circ}_{cell})$ is an extensive property because it depends on the number of moles of electrons $(n)$ transferred,which is related to the amount of substance.
$4$. Electrical work done in a galvanic cell is equal to the decrease in Gibbs energy $(W_{elec} = -\Delta G)$.
$5$. Therefore,the statement that the standard free energy change is an intensive property is false.

(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the reaction is $2 Al + 3 Cu^{2+} \rightarrow 2 Al^{3+} + 3 Cu$.
The number of electrons transferred $(n)$ is $6$ (since $Al \rightarrow Al^{3+} + 3e^-$ and $Cu^{2+} + 2e^- \rightarrow Cu$,multiplied by $2$ and $3$ respectively).
Given $\Delta G^{\circ} = -1158 \ kJ = -1158000 \ J$.
$F = 96500 \ C \ mol^{-1}$.
Substituting the values: $-1158000 = -(6) \times (96500) \times E^{\circ}_{cell}$.
$E^{\circ}_{cell} = \frac{1158000}{6 \times 96500} = \frac{1158000}{579000} = 2 \ V$.

(D) The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
From the Nernst equation at equilibrium,$\Delta G^{\circ} = -nFE^{\circ}_{cell}$,which implies $E^{\circ}_{cell} = \frac{-\Delta G^{\circ}}{nF}$.
Also,$E^{\circ}_{cell} = \frac{RT}{nF} \ln K_{c} = \frac{0.0592}{n} \log_{10} K_{c}$ at $298 \ K$.
Option $D$ is false because the standard cell potential is the difference between the reduction potentials of the cathode and the anode,not the sum.

(D) The given half-reaction is the oxidation of silver: $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} + e^{-}$.
Given $E^{\circ}(Ag^{+}/Ag) = +0.80 \ V$,the standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -0.80 \ V$.
Using the Nernst equation for the oxidation half-cell at $298 \ K$:
$E = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Ag^{+}]$
Here,$n = 1$ and $[Ag^{+}] = 0.01 \ M = 10^{-2} \ M$.
$E = -0.80 - \frac{0.0591}{1} \log(10^{-2})$
$E = -0.80 - 0.0591 \times (-2)$
$E = -0.80 + 0.1182$
$E = -0.6818 \ V \approx -0.6816 \ V$.

(C) The given reaction is the oxidation of $Mg$ to $Mg^{+2}$. The standard reduction potential is $E^{\circ}(Mg^{+2} \mid Mg) = -2.37 \ V$. Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$.
Using the Nernst equation for the oxidation half-cell: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{+2}]$.
Here,$n = 2$ and $[Mg^{+2}] = 0.01 \ M = 10^{-2} \ M$.
Substituting the values: $E_{ox} = 2.37 - \frac{0.0591}{2} \log(10^{-2})$.
$E_{ox} = 2.37 - 0.02955 \times (-2)$.
$E_{ox} = 2.37 + 0.0591 = 2.4291 \ V$.
Rounding to four decimal places,we get $2.4292 \ V$.

(C) The cell reaction is $2 \ Al_{(s)} + 3 \ Ni_{(aq)}^{+2} \rightarrow 2 \ Al_{(aq)}^{+3} + 3 \ Ni_{(s)}$.
Here,$Al$ is oxidized to $Al^{+3}$ (anode) and $Ni^{+2}$ is reduced to $Ni$ (cathode).
The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Given $E^{\circ}_{Ni^{+2}/Ni} = -0.25 \ V$ and $E^{\circ}_{Al^{+3}/Al} = -1.66 \ V$.
$E^{\circ}_{cell} = (-0.25 \ V) - (-1.66 \ V) = -0.25 \ V + 1.66 \ V = +1.41 \ V$.

(C) The standard reduction potential $(E^{\circ})$ values for the given half-reactions are as follows:
$1$. For $2 H_{(aq)}^{+} + 2 e^{-} \rightarrow H_{2_{(g)}}$,$E^{\circ} = 0.00 \ V$.
$2$. For $F_{2_{(g)}} + 2 e^{-} \rightarrow 2 F_{(aq)}^{-}$,$E^{\circ} = +2.87 \ V$.
$3$. For $Li_{(aq)}^{+} + e^{-} \rightarrow Li_{(s)}$,$E^{\circ} = -3.05 \ V$.
$4$. For $Cl_{2_{(g)}} + 2 e^{-} \rightarrow 2 Cl_{(aq)}^{-}$,$E^{\circ} = +1.36 \ V$.
Comparing these values,the reaction involving lithium has the most negative value,which is $-3.05 \ V$. Therefore,it exhibits the minimum standard reduction potential.

(C) The given half-cell reaction is the oxidation of aluminum: $Al_{(s)} \rightarrow Al_{(aq)}^{+3} + 3e^-$.
The standard reduction potential is $E^{\circ}(Al^{+3} \mid Al) = -1.66 \ V$,so the standard oxidation potential is $E^{\circ}_{ox} = +1.66 \ V$.
Using the Nernst equation for the oxidation potential: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log[Al^{+3}]$.
Here,$n = 3$ and $[Al^{+3}] = 0.1 \ M$.
$E_{ox} = 1.66 - \frac{0.0591}{3} \log(0.1)$.
$E_{ox} = 1.66 - 0.0197 \times (-1)$.
$E_{ox} = 1.66 + 0.0197 = 1.6797 \ V \approx +1.679 \ V$.

(A) The standard $emf$ of a cell is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
For a spontaneous cell reaction,the cathode is the electrode with the higher reduction potential.
Here,$E^{\circ}(Sn^{+2} \mid Sn) = -0.14 \ V$ and $E^{\circ}(Fe^{+2} \mid Fe) = -0.44 \ V$.
Since $-0.14 \ V > -0.44 \ V$,the $Sn$ electrode acts as the cathode and the $Fe$ electrode acts as the anode.
Therefore,$E^{\circ}_{cell} = (-0.14 \ V) - (-0.44 \ V) = -0.14 \ V + 0.44 \ V = +0.30 \ V$.

(A) The Nernst equation at $298 \ K$ is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{+2}]}{[Cu^{+2}]}$.
Here,$n = 2$ (number of electrons transferred).
Given $[Zn^{+2}] = 0.1 \ M$ and $[Cu^{+2}] = 0.1 \ M$.
Substituting the values: $E_{cell} = 1.1 - \frac{0.0591}{2} \log \frac{0.1}{0.1}$.
Since $\log(1) = 0$,the equation becomes: $E_{cell} = 1.1 - 0 = 1.1 \ V$.

(D) The given reaction is the reduction of $Cu^{2+}$ to $Cu_{(s)}$: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ with $E^{\circ} = +0.34 \ V$.
The requested reaction is the oxidation of $Cu_{(s)}$ to $Cu^{2+}_{(aq)}$: $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$.
This is the reverse of the reduction reaction multiplied by a factor of $2$.
Since the standard electrode potential $(E^{\circ})$ is an intensive property,it does not depend on the stoichiometric coefficients of the reaction.
Therefore,reversing the reaction changes the sign of the potential,but multiplying the coefficients does not change the value of $E^{\circ}$.
Thus,the potential for the oxidation reaction $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}$ is $-0.34 \ V$.
The potential for $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$ remains $-0.34 \ V$.

(B) In a galvanic cell represented as $Anode | Anode_{electrolyte} || Cathode_{electrolyte} | Cathode$,the oxidation occurs at the anode and reduction occurs at the cathode.
For the given cell $A_{(s)} | A_{(aq)}^{+2} || B_{(aq)}^{+} | B_{(s)}$:
Anode reaction (Oxidation): $A_{(s)} \rightarrow A_{(aq)}^{+2} + 2e^-$.
Cathode reaction (Reduction): $B_{(aq)}^{+} + e^- \rightarrow B_{(s)}$.
To balance the electrons,multiply the cathode reaction by $2$: $2 B_{(aq)}^{+} + 2e^- \rightarrow 2 B_{(s)}$.
Adding both half-reactions: $A_{(s)} + 2 B_{(aq)}^{+} \rightarrow A_{(aq)}^{+2} + 2 B_{(s)}$.
Thus,the correct option is $B$.

(B) In a galvanic cell represented as $Anode | Anode \text{ } electrolyte || Cathode \text{ } electrolyte | Cathode$,the oxidation occurs at the anode and reduction occurs at the cathode.
Given the cell notation: $A_{(s)} | A^{+}_{(1 \ M)} || B^{+2}_{(1 \ M)} | B_{(s)}$.
At the anode (oxidation): $A_{(s)} \rightarrow A^{+}_{(aq)} + e^-$.
At the cathode (reduction): $B^{+2}_{(aq)} + 2e^- \rightarrow B_{(s)}$.
To balance the electrons,multiply the anode reaction by $2$: $2A_{(s)} \rightarrow 2A^{+}_{(aq)} + 2e^-$.
Adding the two half-reactions: $2A_{(s)} + B^{+2}_{(aq)} \rightarrow 2A^{+}_{(aq)} + B_{(s)}$.
Since the $EMF$ is positive,this reaction is spontaneous.

(D) The cell reaction is: $Cd_{(s)} + Cu^{2+}_{(aq)} \rightarrow Cd^{2+}_{(aq)} + Cu_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$,$E^{\circ}_{cell} = 0.74 \ V$.
Given that both concentrations decrease by $10$ times,the ratio $\frac{[Cd^{2+}]}{[Cu^{2+}]}$ remains $\frac{0.1}{0.1} = 1$.
Therefore,$E_{cell} = 0.74 - \frac{0.0591}{2} \log(1)$.
Since $\log(1) = 0$,$E_{cell} = 0.74 \ V$.

(A) The cell reaction is: $Ni_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + 2Ag_{(s)}$.
Applying the Nernst equation at $298 \ K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Ni^{2+}]}{[Ag^{+}]^2} = \frac{0.01}{(0.01)^2} = \frac{0.01}{0.0001} = 100$.
Substituting the values: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log(100)$.
$E_{cell} = E^{\circ}_{cell} - 0.0296 \times 2 = E^{\circ}_{cell} - 0.0592 \ V$.
Thus,the cell emf is less than $E^{\circ}_{cell}$ by $0.0592 \ V$.

(A) The given reaction is the oxidation half-reaction: $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^-$.
The standard reduction potential is $E^{\circ}(Zn^{+2} \mid Zn) = -0.76 \ V$.
Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -(-0.76 \ V) = +0.76 \ V$.
Using the Nernst equation for the oxidation half-cell:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log([Zn^{+2}])$.
Here,$n = 2$ and $[Zn^{+2}] = 0.01 \ M = 10^{-2} \ M$.
$E_{ox} = 0.76 - \frac{0.0591}{2} \log(10^{-2})$.
$E_{ox} = 0.76 - 0.02955 \times (-2)$.
$E_{ox} = 0.76 + 0.0591 = 0.8191 \ V \approx +0.8192 \ V$.

(C) The standard Gibbs energy change $\Delta G^{\circ}$ is related to the standard cell potential $E^{\circ}_{\text{cell}}$ by the equation: $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$.
In the given reaction,$Zn$ is oxidized to $Zn^{2+}$ $(n=2)$ and $Ni^{2+}$ is reduced to $Ni$ $(n=2)$,so the number of electrons transferred $n = 2$.
The Faraday constant $F \approx 96500 \ C \ mol^{-1}$.
Given $E^{\circ}_{\text{cell}} = 0.5 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \ C \ mol^{-1} \times 0.5 \ V = -96500 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -96.5 \ kJ \ mol^{-1}$.

(C) reaction is spontaneous if the standard cell potential $(E^{\circ}_{cell})$ is positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
For option $(C)$: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.
Here,$Zn$ is oxidized (anode) and $Cu^{2+}$ is reduced (cathode).
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Zn^{2+}/Zn} = 0.34 \ V - (-0.76 \ V) = +1.10 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.

(D) The given reaction is the oxidation of copper: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$.
The standard reduction potential is $E^{\circ}(Cu^{2+} \mid Cu) = +0.34 \ V$.
The standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -0.34 \ V$.
Using the Nernst equation for the oxidation half-cell:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Cu^{2+}]$.
Here,$n = 2$ and $[Cu^{2+}] = 0.1 \ M$.
$E_{ox} = -0.34 - \frac{0.0591}{2} \log(0.1)$.
$E_{ox} = -0.34 - 0.02955 \times (-1)$.
$E_{ox} = -0.34 + 0.02955 = -0.31045 \ V \approx -0.3104 \ V$.

(B) The relationship between the standard $EMF$ of the cell $(E^{\circ}_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_c$
Here,the number of electrons transferred $(n)$ is $2$ as $A \rightarrow A^{2+} + 2e^-$ and $B^{2+} + 2e^- \rightarrow B$.
Given $K_c = 10^4$ and $n = 2$.
Substituting the values:
$E^{\circ}_{cell} = \frac{0.0591}{2} \log(10^4)$
$E^{\circ}_{cell} = \frac{0.0591}{2} \times 4$
$E^{\circ}_{cell} = 0.0591 \times 2 = 0.1182 \ V$.
Rounding to the nearest option,we get $0.1184 \ V$.

(B) In a galvanic cell,the electrode with the higher reduction potential acts as the cathode,and the one with the lower reduction potential acts as the anode.
Given $E^{\circ}(Cu^{2+}/Cu) = +0.34 \ V$ and $E^{\circ}(H^{+}/H_2) = 0.00 \ V$.
Since $0.34 \ V > 0.00 \ V$,the copper electrode acts as the cathode (reduction) and the standard hydrogen electrode $(SHE)$ acts as the anode (oxidation).
Anode reaction: $H_{2(g)} \longrightarrow 2H^{+}_{(aq)} + 2e^{-}$
Cathode reaction: $Cu^{2+}_{(aq)} + 2e^{-} \longrightarrow Cu_{(s)}$
Adding these two half-reactions gives the net cell reaction: $H_{2(g)} + Cu^{2+}_{(aq)} \longrightarrow 2H^{+}_{(aq)} + Cu_{(s)}$.

(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} \rightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.
Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Zn^{+2}]}{[Ag^{+1}]^2}$.
Initially,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 1 \ M$,so $Q_1 = \frac{1}{1^2} = 1$. Thus,$E_1 = E^0_{cell} - \frac{0.0592}{2} \log(1) = E^0_{cell}$.
Finally,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 0.1 \ M$,so $Q_2 = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.
$E_2 = E^0_{cell} - \frac{0.0592}{2} \log(100) = E^0_{cell} - 0.0296 \times 2 = E^0_{cell} - 0.0592 \ V$.
The change in $emf$ is $E_2 - E_1 = -0.0592 \ V$,which means the $emf$ decreases by $0.0592 \ V$.

(B) The reduction half-reaction for the hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
According to the Nernst equation at $298 \ K$:
$E_{red} = E^\circ_{red} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^+]^2}$.
Given $E^\circ_{red} = 0 \ V$,$P_{H_2} = 1 \ atm$,$n = 2$,and $pH = 1$ (which means $[H^+] = 10^{-1} \ M$):
$E_{red} = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-1})^2}$.
$E_{red} = -0.02955 \times \log(10^2) = -0.02955 \times 2 = -0.0591 \ V$.
Rounding to the nearest option,the value is $-0.0592 \ V$.

(B) The electrolysis of fused $NaCl$ involves the reaction at the anode: $2Cl^- \rightarrow Cl_2(g) + 2e^-$.
According to the stoichiometry,$1 \ mol$ of $Cl_2$ gas is produced by $2 \ mol$ of electrons.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$.
Therefore,$22.4 \ dm^3$ of $Cl_2$ requires $2 \times 96500 \ C$ of electricity.
For $0.224 \ dm^3$ of $Cl_2$,the quantity of electricity required is:
$Q = \frac{2 \times 96500 \times 0.224}{22.4} \ C$.
$Q = 2 \times 96500 \times 0.01 \ C = 1930 \ C$.

(D) $1$. In the electrolysis of molten $NaCl$,the reactions are: At cathode: $Na^+ + e^- \rightarrow Na_{(s)}$; At anode: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
$2$. In the electrolysis of aqueous $NaCl$ (brine),the reactions are: At cathode: $2H_2O + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$; At anode: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
$3$. Comparing both processes,$Cl_{2(g)}$ gas is liberated at the anode in both cases.

(A) In a hydrogen-oxygen fuel cell,the following reactions occur at the electrodes:
At the anode: $2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-$
At the cathode: $O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$
In the anode reaction,$H_2$ undergoes oxidation (loss of electrons),which means it acts as a reducing agent.
Therefore,$H_2$ is the species that acts as the reducing agent.

(C) In a dry cell (Leclanché cell),the anode is the zinc container,where oxidation occurs: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$.
The cathode (positive electrode) is a graphite rod surrounded by powdered $MnO_2$ and carbon.
At the cathode,the reduction reaction occurs: $MnO_{2\text{(s)}} + NH_4^+{_{\text{(aq)}}} + e^{-} \rightarrow MnO(OH)_{\text{(s)}} + NH_{3\text{(g)}}$.
This reaction involves the reduction of $Mn$ from the $+4$ oxidation state in $MnO_2$ to the $+3$ oxidation state in $MnO(OH)$,which is often represented as $Mn_2O_3$.
Thus,$MnO_{2_{(s)}}$ is reduced to $Mn_2O_{3_{(s)}}$.

(B) During the discharging of a lead storage battery,the positive electrode $(PbO_2)$ is reduced to $PbSO_4$.
During the recharging process,the chemical reactions are reversed.
At the positive electrode (anode during recharging),$PbSO_4$ is oxidised to $PbO_2$ according to the following reaction:
$PbSO_4(s) + 2H_2O(l) \rightarrow PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^-$.

(B) The molar conductivity $(\Lambda_m)$ of a solution is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution of volume $V \ mL$.
The relationship is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$,where $k$ is the conductivity in $S \ cm^{-1}$ and $M$ is the molarity in $mol \ L^{-1}$.
Therefore,option $B$ is the correct relationship.

(A) In solid sodium chloride $(NaCl)$,the ions are held together by strong electrostatic forces in a rigid crystal lattice.
Since the ions are not free to move,solid $NaCl$ cannot conduct electricity.
In contrast,aqueous $KCl$ contains free ions,while graphite and copper metal contain free electrons,allowing them to conduct electricity.

(D) The chemical formula for aluminium sulphate is $Al_2(SO_4)_3$.
According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution (zero concentration) is given by:
$\Lambda_m^0 (Al_2(SO_4)_3) = 2 \lambda_m^0 (Al^{3+}) + 3 \lambda_m^0 (SO_4^{2-})$
Given:
$\lambda_m^0 (Al^{3+}) = 189 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\lambda_m^0 (SO_4^{2-}) = 50.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Substituting the values:
$\Lambda_m^0 = 2(189) + 3(50.1)$
$\Lambda_m^0 = 378 + 150.3 = 528.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Thus,the correct option is $D$.

(C) The cell constant $(G^*)$ is defined by the formula: $G^* = \kappa \times R$,where $\kappa$ is the conductivity and $R$ is the resistance.
Given:
Conductivity $(\kappa)$ = $0.001262 \ \Omega^{-1} \ cm^{-1}$
Resistance $(R)$ = $1440 \ \Omega$
Calculation:
$G^* = 0.001262 \ \Omega^{-1} \ cm^{-1} \times 1440 \ \Omega$
$G^* = 1.81728 \ cm^{-1}$
Rounding to three decimal places,we get $1.817 \ cm^{-1}$.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Rearranging for resistivity,we get $\rho = R \frac{A}{l}$.
The unit of resistance $R$ is $\Omega$ (ohm),the unit of area $A$ is $m^2$,and the unit of length $l$ is $m$.
Substituting these units,the unit of resistivity $\rho = \Omega \times \frac{m^2}{m} = \Omega \ m$.
Therefore,the $SI$ unit of resistivity is $\Omega \ m$.

(B) $Freon-12$ is the common name for the chemical compound $CCl_2F_2$.
Its $IUPAC$ name is $Dichlorodifluoromethane$.
It is widely used as a refrigerant in air conditioning and refrigeration systems.

(A) The principle of $12$ Principles of Green Chemistry that suggests avoiding unnecessary derivatization (such as protection and deprotection steps) is known as "Reduce derivatives".
This principle aims to minimize the use of temporary groups,as they require additional reagents and generate waste,thereby increasing the atom economy of the process.

(A) $BHC$ (Benzene Hexachloride),also known as Gammexane or Lindane,is used as an effective insecticide and is often considered as an alternative to $DDT$ (Dichlorodiphenyltrichloroethane) in various agricultural applications.