Electrochemical cells Questions in English

One of the most successful fuel cells is the hydrogen-oxygen fuel cell.
Principle: Chemical energy of hydrogen fuel is directly converted into electrical energy. It uses the reaction of hydrogen with oxygen to form water.
Construction: The fuel cell consists of two porous carbon electrodes. Hydrogen gas is passed through the anode and oxygen gas is passed through the cathode. An aqueous solution of sodium hydroxide is used as the electrolyte. Catalysts like finely divided platinum or palladium are incorporated into the electrodes to increase the rate of electrode reactions.
Working and reaction: Hydrogen and oxygen are bubbled through the porous carbon electrodes into the concentrated aqueous sodium hydroxide solution. Oxidation of $H_2$ occurs at the anode and reduction of $O_2$ occurs at the cathode.
Cathode reaction: $O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^-_{(aq)}$
Anode reaction: $2H_{2(g)} + 4OH^-_{(aq)} \rightarrow 4H_2O_{(l)} + 4e^-$
Overall cell reaction: $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
Thus,water is produced by the direct reaction of $H_2$ and $O_2$,converting chemical energy directly into electrical energy.

(N/A) Anodic reaction: $2H_{2(g)} + 4OH^{-}_{(aq)} \rightarrow 4H_2O_{(l)} + 4e^-$
Cathodic reaction: $O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^{-}_{(aq)}$
Overall reaction: $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$

(D) In a fuel cell,the electrochemical reactions (oxidation of fuel and reduction of oxidant) occur at the electrode-electrolyte interface.
Finely divided $Pt$ (platinum) and $Pd$ (palladium) are used as electrode materials because they act as efficient catalysts for these reactions.
The finely divided state increases the effective surface area of the electrodes,which significantly enhances the rate of the electrochemical reactions.
Therefore,both the catalytic property and the increased surface area are essential for the efficient functioning of the fuel cell.
Thus,the correct option is $D$.

(N/A) The advantages of fuel cells are as follows:
$1$. High Efficiency: Fuel cells convert the chemical energy of fuels like $H_2$ and $CO$ directly into electrical energy with high efficiency (typically $70-80\%$),unlike thermal power plants which are limited by Carnot cycle efficiency.
$2$. Pollution-Free: Since the byproduct of hydrogen-oxygen fuel cells is water $(H_2O)$,they are environmentally friendly and do not produce harmful pollutants.
$3$. Continuous Supply: As long as the reactants ($H_2$ and $O_2$) are supplied,the cell will continue to produce electricity.
$4$. Lightweight and Compact: They are ideal for space applications due to their high power-to-weight ratio.

(A) In a hydrogen-oxygen fuel cell,the following reactions occur at the electrodes:
At the anode: $2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-$. Here,$H_2$ undergoes oxidation.
At the cathode: $O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$. Here,$O_2$ undergoes reduction.
Therefore,$H_2$ is oxidized and $O_2$ is reduced.

(A) $1.$ True $(T)$: $A$ fuel cell is a type of galvanic cell that converts the chemical energy of a fuel (like $H_2$) and an oxidant (like $O_2$) directly into electrical energy.
$2.$ False $(F)$: Coal is not used as a fuel in fuel cells. Typically,gases like $H_2$,$CH_4$,or $CH_3OH$ are used as fuels in fuel cells.

(T, T, T) $1.$ $T$: $H_2$ is the most common fuel used in hydrogen-oxygen fuel cells.
$2.$ $T$: Thermal power plants burn fossil fuels,releasing greenhouse gases and pollutants,whereas fuel cells produce electricity through electrochemical reactions with minimal emissions.
$3.$ $T$: Fuel cells are considered pollution-free because the only byproduct of a hydrogen-oxygen fuel cell is water $(H_2O)$.

(B) The relationship between Gibbs energy change and cell potential is given by $\Delta_r G = -nFE_{cell}$.
When a chemical reaction reaches equilibrium,the net reaction stops,meaning no further work can be done by the cell.
At this state,the cell potential $E_{cell}$ becomes $0$,and consequently,the Gibbs energy change $\Delta_r G$ also becomes $0$.

(A) In the symbolic representation of a galvanic cell,the oxidation half-reaction is written on the left side and the reduction half-reaction on the right side.
In the given reaction,the oxidation of $Cu$ to $Cu^{2+}$ and the reduction of $Ag^{+}$ to $Ag$ occurs.
The two vertical lines represent the salt bridge placed between these two half-cells.
The complete representation of the cell is: $Cu | Cu^{2+} || Ag^{+} | Ag$

(N/A) The oxidation reactions at the anode are:
$(i)$ $Cl_{(aq)}^{-} \rightarrow \frac{1}{2} Cl_{2(g)} + e^{-}$; $E^{\circ} = 1.36 \ V$
$(ii)$ $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H_{(aq)}^{+} + 4e^{-}$; $E^{\circ} = 1.23 \ V$
Although the standard oxidation potential of water $(1.23 \ V)$ is lower than that of $Cl^{-}$ $(1.36 \ V)$,which theoretically suggests water should be oxidized,the oxidation of $Cl^{-}$ is observed.
This occurs due to the 'overpotential' of oxygen. The evolution of oxygen gas at the anode requires an extra potential (overpotential) beyond the standard value,making the effective potential for water oxidation significantly higher than $1.36 \ V$. Consequently,$Cl^{-}$ ions are preferentially oxidized to $Cl_{2}$ gas.

(A) In the given electrochemical cell (Daniell cell),the $Zn$ electrode acts as the anode (negative terminal) and the $Cu$ electrode acts as the cathode (positive terminal).
The electrode $A$ is connected to the $Zn$ electrode of the electrochemical cell. Since the $Zn$ electrode is the negative terminal,the electrode $A$ becomes the positive terminal (anode) of the electrolytic cell.
The electrode $B$ is connected to the $Cu$ electrode of the electrochemical cell. Since the $Cu$ electrode is the positive terminal,the electrode $B$ becomes the negative terminal (cathode) of the electrolytic cell.
Therefore,electrode $A$ is positive and electrode $B$ is negative.

(B) When an external opposing potential equal to the cell potential $(1.1 \ V)$ is applied,the potential difference becomes zero.
As a result,the chemical reaction in the cell stops,and no current flows through the cell.

(N/A) The cell potential of a mercury cell remains constant throughout its useful life because the overall cell reaction does not involve any ions whose concentration could change.
In a mercury cell,$Zn(Hg)$ is used as the anode and $HgO$ with carbon paste is used as the cathode. The electrode reactions are as follows:
Cathode: $HgO_{(s)} + H_2O_{(l)} + 2e^- \rightarrow Hg_{(l)} + 2OH^-_{(aq)}$
Anode: $Zn(Hg) + 2OH^-_{(aq)} \rightarrow ZnO_{(s)} + H_2O_{(l)} + 2e^-$
Overall cell reaction: $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$
Since the overall reaction involves only solids and liquids,the concentration of the species remains constant,resulting in a steady cell potential of $1.35 \ V$.

(N/A) In the given cell representation,the left side represents the anode and the right side represents the cathode.
Oxidation reaction at the anode: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-$
Reduction reaction at the cathode: $Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$
At the anode,$Cu$ undergoes oxidation by losing electrons,and at the cathode,$Cl_2$ undergoes reduction by gaining electrons.

(B) Primary batteries contain a limited amount of reactants and are discharged when the reactants are consumed.
Secondary batteries can be recharged,but the process is time-consuming.
Fuel cells have the advantage of running continuously as long as the reactants (e.g.,$H_2$ and $O_2$) are supplied to them and the products are removed continuously.

(N/A) The cell reactions during the discharge of a lead storage battery are as follows:
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Overall reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
During discharge,$H_2SO_4$ is consumed and water is produced,which leads to a decrease in the density of the electrolyte.

(N/A) $(i)$ Cell $'B'$ will act as an electrolytic cell because it has a lower $emf$ $(1.1 \ V < 2 \ V)$.
$\therefore$ The electrode reactions in cell $'B'$ will be:
Reduction at cathode: $Zn^{2+} + 2e^{-} \rightarrow Zn$
Oxidation at anode: $Cu \rightarrow Cu^{2+} + 2e^{-}$
$(ii)$ Now,cell $'B'$ acts as a galvanic cell because it has a higher $emf$ $(1.1 \ V > 0.5 \ V)$ and will push electrons into cell $'A'$.
The electrode reactions will be:
At anode: $Zn \rightarrow Zn^{2+} + 2e^{-}$
At cathode: $Cu^{2+} + 2e^{-} \rightarrow Cu$

(N/A) $(i)$ Electrons move from the $Zn$ electrode to the $Ag$ electrode.
$(ii)$ The silver plate acts as the cathode.
$(iii)$ The cell will stop functioning because the electrical circuit is broken.
$(iv)$ The cell stops functioning when equilibrium is attained,i.e.,$E_{cell} = 0$.
$(v)$ As the cell functions,the concentration of $Zn^{2+}$ ions increases due to the oxidation of $Zn$ at the anode,and the concentration of $Ag^{+}$ ions decreases due to the reduction of $Ag^{+}$ at the cathode.
$(vi)$ When the cell becomes 'dead' $(E_{cell} = 0)$,the system is at equilibrium,so the concentrations of $Zn^{2+}$ and $Ag^{+}$ ions will no longer change.

(A) The cell reaction is $Cu_{(s)} + Sn^{2+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Sn_{(s)}$.
The standard cell potential is $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Sn^{2+}|Sn} - E^{\circ}_{Cu^{2+}|Cu}$.
$E^{\circ}_{cell} = -0.16 \, V - 0.34 \, V = -0.50 \, V$.
The standard Gibbs energy change is $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,$n = 2$ (number of electrons transferred).
$\Delta G^{\circ} = -2 \times 96500 \, C \, mol^{-1} \times (-0.50 \, V) = 96500 \, J \, mol^{-1}$.
Since the concentrations are $[Cu^{2+}] = [Sn^{2+}] = 1 \, M$,the reaction quotient $Q = \frac{[Cu^{2+}]}{[Sn^{2+}]} = 1$.
Using the Nernst equation,$\Delta G = \Delta G^{\circ} + RT \ln Q$.
Since $\ln(1) = 0$,$\Delta G = \Delta G^{\circ} = 96500 \, J \, mol^{-1}$.

(A) The standard cell potential is calculated as:
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.34 \ V - (-0.76 \ V) = 1.10 \ V$.
If $E_{\text{ext}} < 1.1 \ V$:
Electrons flow from $Zn$ to $Cu$. $Zn$ acts as the anode (dissolves) and $Cu$ acts as the cathode (deposits).
If $E_{\text{ext}} = 1.1 \ V$:
No current flows,and the chemical reaction stops.
If $E_{\text{ext}} > 1.1 \ V$:
The external potential overcomes the cell potential. The cell acts as an electrolytic cell. $Cu$ acts as the anode (dissolves) and $Zn$ acts as the cathode (deposits). Electrons flow from $Cu$ to $Zn$.
Therefore,option $A$ is the incorrect statement.

(C) The most common type of fuel cell is the hydrogen-oxygen fuel cell.
In this cell,hydrogen gas $(H_{2(g)})$ and oxygen gas $(O_{2(g)})$ are the reactants $(R)$.
The overall cell reaction is:
$2H_{2(g)} + O_{2(g)} \longrightarrow 2H_{2}O_{(l)}$
Thus,the product $(P)$ is water $(H_{2}O_{(l)})$.

(C) The net cell reaction is $Zn_{(s)} + Ag_{2}O_{(s)} \rightarrow ZnO_{(s)} + 2Ag_{(s)}$.
The standard Gibbs free energy change for the reaction is given by $\Delta G^{o} = \Delta G_{f}^{o}(ZnO) - \Delta G_{f}^{o}(Ag_{2}O)$.
Substituting the values: $\Delta G^{o} = -318.3 - (-11.21) = -307.09 \ kJ \ mol^{-1} = -307.09 \times 10^{3} \ J \ mol^{-1}$.
Using the relation $\Delta G^{o} = -n F E^{o}_{cell}$,where $n = 2$ (number of electrons transferred) and $F = 96500 \ C \ mol^{-1}$.
$-307.09 \times 10^{3} = -2 \times 96500 \times E^{o}_{cell}$.
$E^{o}_{cell} = \frac{307.09 \times 10^{3}}{2 \times 96500} = 1.591 \ V$.

(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (OH^{-})$
Given values:
$\lambda_{m}^{o} (NaOH) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (OH^{-}) = 248 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$\lambda_{m}^{o} (NaCl) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (Cl^{-}) = 126 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$\lambda_{m}^{o} (BaCl_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (Cl^{-}) = 280 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
To obtain $\lambda_{m}^{o} (Ba(OH)_{2})$,we perform the operation:
$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (BaCl_{2}) + 2\lambda_{m}^{o} (NaOH) - 2\lambda_{m}^{o} (NaCl)$
Substituting the values:
$\lambda_{m}^{o} (Ba(OH)_{2}) = 280 \times 10^{-4} + 2(248 \times 10^{-4}) - 2(126 \times 10^{-4})$
$= (280 + 496 - 252) \times 10^{-4} \ S \ m^{2} \ mol^{-1}$
$= 524 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$

(D) The reaction is $2 Fe^{3+} + 2 I^{-} \rightarrow 2 Fe^{2+} + I_{2}$.
First,calculate $E^{\circ}_{Fe^{3+}/Fe^{2+}}$ using the given Latimer diagram data:
$n_1 E_1^{\circ} + n_2 E_2^{\circ} = n_3 E_3^{\circ}$
$1 \times E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2 \times E^{\circ}_{Fe^{2+}/Fe} = 3 \times E^{\circ}_{Fe^{3+}/Fe}$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2(-0.440) = 3(-0.036)$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} = -0.108 + 0.880 = 0.772 \ V$.
Now,calculate the cell potential $E^{\circ}_{cell}$:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_{2}/2I^{-}}$
$E^{\circ}_{cell} = 0.772 - 0.539 = 0.233 \ V$.
The standard Gibbs free energy change is $\Delta_{r} G^{\circ} = -n F E^{\circ}_{cell}$.
Here,$n = 2$ (moles of electrons transferred).
$\Delta_{r} G^{\circ} = -2 \times 96500 \times 0.233 = -44969 \ J \ mol^{-1} = -44.969 \ kJ \ mol^{-1}$.
The magnitude is approximately $45 \ kJ \ mol^{-1}$.

(A) $PbO_2$ is an amphoteric oxide and a strong oxidizing agent.
It acts as the cathode material in lead storage batteries.

(B) The cell reaction is the sum of the cathode and anode reactions:
Cathode: $Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^- ; E^{\circ} = 0.344 \, V$
Anode: $Zn \rightarrow Zn^{2+} + 2e^- ; E^{\circ} = +0.760 \, V$
Overall $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.344 - (-0.760) = 1.104 \, V$
Using the formula $\Delta G^{\circ} = -nFE_{cell}^{\circ}$ where $n = 2$:
$\Delta G^{\circ} = -2 \times 96,500 \times 1.104 \, J \, mol^{-1} = -213,072 \, J \, mol^{-1}$
Converting to $kJ \, mol^{-1}$: $\Delta G^{\circ} = -213.072 \, kJ \, mol^{-1}$

(C) The standard reduction potentials $(SRP)$ are given as: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$,$E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$,and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$.
$A$ metal with a lower $SRP$ value can displace a metal with a higher $SRP$ value from its salt solution.
Comparing the values,the reactivity order is $Zn > Fe > Cu > Ag$.
In option $C$,$Ag$ is less reactive than $Cu$ $(E^{\circ}_{Ag^{+}/Ag} > E^{\circ}_{Cu^{2+}/Cu})$,therefore $Ag$ cannot displace $Cu$ from $CuSO_4$ solution.
Thus,the reaction $2CuSO_{4(aq)} + 2Ag_{(s)} \rightarrow 2Cu_{(s)} + Ag_2SO_{4(aq)}$ is not spontaneous and cannot occur.

(D) The correct option is $D$. When a copper rod is dipped into an aqueous solution of $AgNO_{3}$,a displacement reaction occurs because copper is more reactive than silver.
$Cu_{(s)} + 2AgNO_{3(aq)} \longrightarrow Cu(NO_{3})_{2(aq)} + 2Ag_{(s)}$
In this reaction,$Cu$ is oxidized to $Cu^{2+}$ ions,which impart a blue color to the solution.
The half-cell reactions are:
$Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \, V$
$Ag^{+}_{(aq)} + e^{-} \longrightarrow Ag_{(s)}; E^{\circ} = 0.80 \, V$
The overall cell potential is positive,indicating the reaction is spontaneous.

(C) The cell reaction is $3 Fe^{2+}_{(aq)} + 2 Cr_{(s)} \rightleftharpoons 2 Cr^{3+}_{(aq)} + 3 Fe_{(s)}$.
Here,$Cr$ is oxidized to $Cr^{3+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).
The standard cell potential is $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{Fe^{2+}/Fe}^{\circ} - E_{Cr^{3+}/Cr}^{\circ}$.
$E_{cell}^{\circ} = -0.44 \, V - (-0.74 \, V) = 0.30 \, V$.
The number of electrons transferred $(n)$ is $6$.
The standard free energy change is given by $\Delta G^{\circ} = -n F E_{cell}^{\circ}$.
$\Delta G^{\circ} = -6 \times 96500 \, C \times 0.30 \, V = -173,700 \, J$.

(B) The cell reaction for a Daniell cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
Here,$n = 2$ (number of electrons transferred).
The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34\, V - (-0.76\, V) = 1.1\, V$.
The standard Gibbs free energy change is given by: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500\, C\, mol^{-1} \times 1.1\, V = -212300\, J\, mol^{-1} = -212.3\, kJ\, mol^{-1} \approx -213\, kJ\, mol^{-1}$.
Thus,the emf is $1.1\, V$ and the free energy change is $-213\, kJ\, mol^{-1}$.

(C) The reaction in a $Daniell$ cell is: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.
For $1 \, mole$ of $Zn$ oxidation,$n = 2$ electrons are transferred.
For $2 \, moles$ of $Zn$ oxidation,the number of electrons transferred is $n = 2 \times 2 = 4$.
Given $E^{\circ}_{cell} = 1.1 \, V$ and $F = 96500 \, C \, mol^{-1}$.
The formula for standard Gibbs free energy change is $\Delta G^{\circ} = -n F E^{\circ}_{cell}$.
Substituting the values: $\Delta G^{\circ} = -4 \times 96500 \, C \, mol^{-1} \times 1.1 \, V$.
$\Delta G^{\circ} = -424600 \, J \, mol^{-1} = -424.6 \, kJ \, mol^{-1}$.

(B)
According to the electrochemical series,a metal with a higher reduction potential (less reactive) cannot displace a metal with a lower reduction potential (more reactive) from its salt solution.
$Cu^{2+} + 2e^- \rightarrow Cu$ $(E^o = +0.34 \ V)$
$Sn^{2+} + 2e^- \rightarrow Sn$ $(E^o = -0.14 \ V)$
Since $Sn$ has a more negative reduction potential than $Cu$,it is more reactive and can displace $Cu$ from $CuSO_4$ solution.
The reaction is: $Sn(s) + CuSO_4(aq) \rightarrow SnSO_4(aq) + Cu(s)$.

(B) .
$A$ concentrated solution of lead nitrate,$Pb(NO_3)_2$,can be stored in a copper vessel because copper is less reactive than lead in the electrochemical series.
Therefore,copper cannot displace lead from its nitrate solution.
In contrast,iron $(Fe)$,zinc $(Zn)$,and magnesium $(Mg)$ are all more reactive than lead $(Pb)$ and would undergo a displacement reaction,making them unsuitable for storage.

(C) The cell reaction is $Zn_{(s)} + Sn^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Sn_{(s)}$.
For this reaction,the number of electrons transferred is $n = 2$.
The relationship between standard cell potential and equilibrium constant is given by $E_{cell}^0 = \frac{0.059}{n} \log_{10} K_{eq}$.
Substituting the values: $E_{cell}^0 = \frac{0.059}{2} \log_{10} (1 \times 10^{20}) = \frac{0.059 \times 20}{2} = 0.59 \ V$.
We know that $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = E_{Sn^{2+}/Sn}^0 - E_{Zn^{2+}/Zn}^0$.
$0.59 \ V = E_{Sn^{2+}/Sn}^0 - (-0.76 \ V)$.
$E_{Sn^{2+}/Sn}^0 = 0.59 - 0.76 = -0.17 \ V$.
The standard electrode potential for $Sn/Sn^{2+}$ is the negative of $E_{Sn^{2+}/Sn}^0$,so $E_{Sn/Sn^{2+}}^0 = -(-0.17 \ V) = 0.17 \ V$.
$0.17 \ V = 17 \times 10^{-2} \ V$.
Thus,the magnitude is $17$.

(B) The given cell reaction is: $\frac{1}{2} H_{2(g)} + AgCl_{(s)} \rightleftharpoons H^{+}_{(aq)} + Cl^{-}_{(aq)} + Ag_{(s)}$.
At the anode,oxidation occurs: $\frac{1}{2} H_{2(g)} \rightarrow H^{+}_{(aq)} + e^-$.
At the cathode,reduction occurs: $AgCl_{(s)} + e^- \rightarrow Ag_{(s)} + Cl^{-}_{(aq)}$.
Combining these,we get the overall reaction. This corresponds to a cell with a hydrogen electrode $(Pt \mid H_2)$ and a silver-silver chloride electrode $(AgCl \mid Ag)$ in an $HCl$ solution.
Therefore,the correct cell representation is $Pt \mid H_{2(g)} \mid HCl_{(aq)} \mid AgCl_{(s)} \mid Ag_{(s)}$.

(C) The equation $\Delta_r G = -nFE_{\text{cell}}$ relates the Gibbs energy change to the number of moles of electrons $(n)$ transferred,the Faraday constant $(F)$,and the cell potential $(E_{\text{cell}})$.
Since $\Delta_r G$ is directly proportional to $n$,the value of $\Delta_r G$ depends on $n$. Thus,Assertion $A$ is true.
$E_{\text{cell}}$ is an intensive property because it does not depend on the amount of matter,whereas $\Delta_r G$ is an extensive property because it depends on the amount of matter (number of moles $n$). Thus,Reason $R$ is true.
However,the fact that $E_{\text{cell}}$ is intensive and $\Delta_r G$ is extensive explains why $\Delta_r G$ is extensive,but it is not the direct reason why $\Delta_r G$ depends on $n$ in the equation. The dependence on $n$ is a mathematical consequence of the stoichiometry of the cell reaction. Therefore,$R$ is not the correct explanation of $A$.

(A) In battery industries,various metals are used as electrodes or components.
$Mn$ (Manganese) is used in Leclanche cells (dry cells).
$Ni$ (Nickel) is used in Nickel-Cadmium batteries.
$Cd$ (Cadmium) is used in Nickel-Cadmium batteries.
Therefore,the metals $Mn$,$Ni$,and $Cd$ are employed in battery industries.
This corresponds to options $B$,$C$,and $E$.

(A) The calomel electrode is represented as $Hg | Hg_2Cl_2(s) | Cl^-(aq)$.
It consists of mercury in contact with solid mercurous chloride $(Hg_2Cl_2)$ and a solution of chloride ions $(Cl^-)$.
This structure classifies it as a $Metal - Insoluble Salt - Anion$ electrode.

(C) $1$. Fuel cells,such as the $H_2-O_2$ fuel cell,were used in the Apollo space program,making statement $A$ correct.
$2$. Fuel cells are highly efficient,typically converting $60 \%$ to $70 \%$ of the energy of fuels into electricity,unlike thermal plants which are limited by Carnot efficiency. Statement $B$ is generally considered correct in the context of high efficiency,though $40 \%$ is a lower bound.
$3$. Fuel cells typically use catalysts like finely divided platinum or palladium,not aluminium. Statement $C$ is incorrect.
$4$. The only byproduct of the $H_2-O_2$ fuel cell is water,making it eco-friendly. Statement $D$ is correct.
$5$. $A$ fuel cell is a device that converts chemical energy directly into electrical energy,which is the definition of a galvanic cell. Statement $E$ is correct.
Therefore,statements $A, B, D,$ and $E$ are correct.

(A) The cells commonly used in clocks are mercury cells or zinc-carbon dry cells. In these cells,the cathode reaction involves the reduction of manganese dioxide $(MnO_2)$.
Specifically,in the dry cell,the reaction at the cathode is: $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$. In $MnO(OH)$,the oxidation state of $Mn$ is $+3$.
Therefore,the reaction involves the reduction of $Mn$ from $+4$ to $+3$.

(D) The cell reaction is $M + X^{2-} \rightarrow M^{2+} + X$.
Standard cell potential $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$.
Here,the cathode is $X/X^{2-}$ and the anode is $M/M^{2+}$.
Given $E^{\circ}_{(M^{2+}/M)} = 0.46 \ V$,so $E^{\circ}_{(M/M^{2+})} = -0.46 \ V$.
Given $E^{\circ}_{(X/X^{2-})} = 0.34 \ V$.
$E^{\circ}_{\text{cell}} = 0.34 \ V - (-0.46 \ V) = 0.80 \ V$.
Since $E^{\circ}_{\text{cell}} > 0$,the reaction $M + X^{2-} \rightarrow M^{2+} + X$ is spontaneous.
However,checking the options,if we consider the reverse reaction $M^{2+} + X^{2-} \rightarrow M + X$,it would be non-spontaneous. Given the options provided,$E^{\circ}_{\text{cell}} = 0.80 \ V$ is the correct value.

(A) An electrochemical cell functions as a galvanic cell when the external potential is less than $E_{\text{cell}}^0$.
When an external potential greater than $E_{\text{cell}}^0$ is applied in the opposite direction,the flow of electrons is reversed,and the cell acts as an electrolytic cell.

(A) The given redox reaction is: $\frac{1}{2} H_{2(g)} + AgCl_{(s)} \rightarrow H_{(aq)}^{+} + Cl_{(aq)}^{-} + Ag_{(s)}$
Anodic half-cell reaction (Oxidation):
$\frac{1}{2} H_{2(g)} \rightarrow H_{(aq)}^{+} + e^{-}$
This corresponds to the hydrogen electrode: $Pt|H_{2(g)}|H_{(aq)}^{+}$
Cathodic half-cell reaction (Reduction):
$AgCl_{(s)} + e^{-} \rightarrow Ag_{(s)} + Cl_{(aq)}^{-}$
This corresponds to the silver-silver chloride electrode: $Cl_{(aq)}^{-}|AgCl_{(s)}|Ag$
Combining these,the cell representation is:
$Pt|H_{2(g)}|H_{(aq)}^{+}, Cl_{(aq)}^{-}|AgCl_{(s)}|Ag$
Since $HCl$ or $KCl$ provides the $H^{+}$ and $Cl^{-}$ ions,the representation $Pt|H_{2(g)}|HCl_{(soln.)}|AgCl_{(s)}|Ag$ is correct. Option $A$ represents this cell.

(C) metal will be oxidized by $Cr_2O_7^{2-}$ if the standard reduction potential $(E^{\circ})$ of the metal is less than the $E^{\circ}$ of the $Cr_2O_7^{2-} / Cr^{3+}$ half-cell $(1.33 \ V)$.
Comparing the given values:
$1. E^{\circ}(Fe^{3+}/Fe) = -0.04 \ V < 1.33 \ V$ (Oxidized)
$2. E^{\circ}(Ni^{2+}/Ni) = -0.25 \ V < 1.33 \ V$ (Oxidized)
$3. E^{\circ}(Ag^+/Ag) = 0.80 \ V < 1.33 \ V$ (Oxidized)
$4. E^{\circ}(Au^{3+}/Au) = 1.40 \ V > 1.33 \ V$ (Not oxidized)
Thus,$Fe$,$Ni$,and $Ag$ will be oxidized.
The total number of such metals is $3$.

(C) The correct matching is as follows:
$A$. Leclanche cell: The anode reaction is $Zn \rightarrow Zn^{2+} + 2e^-$. Thus,$A-IV$.
$B$. Ni-Cd cell: This is a rechargeable cell. Thus,$B-III$.
$C$. Fuel cell: It converts the energy of combustion of fuels like $H_2$ directly into electrical energy. Thus,$C-I$.
$D$. Mercury cell: It does not involve any ion in the solution,so the cell potential remains constant during its life. It is used in hearing aids. Thus,$D-II$.
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(C) Intensive properties are independent of the amount of matter present in the system.
$1$. Molar conductivity is an intensive property because it is defined as the conductivity per mole of electrolyte.
$2$. Electromotive force $(EMF)$ is an intensive property because it is a potential difference independent of the size of the cell.
$3$. Resistance $(R)$ is an extensive property because it depends on the dimensions and amount of material.
$4$. Heat capacity $(C)$ is an extensive property because it depends on the total amount of substance.
Therefore,$A$ and $B$ are intensive properties.

(D) $(P) \ E^{\circ}(Fe^{3+}, Fe) = \frac{1(0.77) + 2(-0.44)}{3} = \frac{0.77 - 0.88}{3} = \frac{-0.11}{3} = -0.0366 \ V \approx -0.04 \ V$
$(Q) \ E^{\circ}(4H_{2}O \rightleftharpoons 4H^{+} + 4OH^{-}) = 0.40 - 1.23 = -0.83 \ V$
$(R) \ E^{\circ}(Cu^{2+} + Cu \rightarrow 2Cu^{+}) = E^{\circ}(Cu^{2+}, Cu^{+}) - E^{\circ}(Cu^{+}, Cu)$
Using the relation $2E^{\circ}(Cu^{2+}, Cu) = E^{\circ}(Cu^{2+}, Cu^{+}) + E^{\circ}(Cu^{+}, Cu)$:
$2(0.34) = E^{\circ}(Cu^{2+}, Cu^{+}) + 0.52 \implies E^{\circ}(Cu^{2+}, Cu^{+}) = 0.68 - 0.52 = 0.16 \ V$
$E^{\circ}_{cell} = E^{\circ}(Cu^{2+}, Cu^{+}) - E^{\circ}(Cu^{+}, Cu) = 0.16 - 0.52 = -0.36 \ V \approx -0.4 \ V$
$(S) \ E^{\circ}(Cr^{3+}, Cr^{2+})$
$3E^{\circ}(Cr^{3+}, Cr) = E^{\circ}(Cr^{3+}, Cr^{2+}) + 2E^{\circ}(Cr^{2+}, Cr)$
$3(-0.74) = E^{\circ}(Cr^{3+}, Cr^{2+}) + 2(-0.91)$
$-2.22 = E^{\circ}(Cr^{3+}, Cr^{2+}) - 1.82$
$E^{\circ}(Cr^{3+}, Cr^{2+}) = -0.40 \ V$
Thus,the correct match is $P-3, Q-4, R-2, S-2$ (Note: The provided options suggest $P-3, Q-4, R-1, S-2$ based on standard approximations,but $R$ calculates to $-0.36 \ V$,closest to $-0.4 \ V$). Given the options,the correct choice is $(D)$.

(A) The salt bridge is used to maintain electrical neutrality in the two half-cells and to complete the electrical circuit.
It does not participate in the chemical reaction occurring at the electrodes.
It allows the flow of ions to maintain charge balance but does not completely stop the diffusion of ions.
It is not strictly necessary for the occurrence of a cell reaction,as some cell designs (like certain types of batteries) function without a salt bridge by using porous membranes or other methods to keep electrolytes separate.

(B) In a hydrazine-oxygen fuel cell,the following reactions occur:
Anode: $N_2H_{4(aq)} + 4OH^{-}_{(aq)} \longrightarrow N_{2(g)} + 4H_2O_{(l)} + 4e^-$
Cathode: $O_{2(g)} + 2H_2O_{(l)} + 4e^- \longrightarrow 4OH^{-}_{(aq)}$
Statement $A$ is correct as it describes the anodic oxidation of hydrazine.
Statement $C$ is correct as it describes the cathodic reduction of oxygen to hydroxide ions.
Statements $B$ and $D$ are incorrect as they do not describe the standard mechanism of this fuel cell.

(A) The relationship between standard Gibbs free energy change and standard cell potential is given by $\Delta G^0 = -nFE^0_{cell}$.
For a cell to have the most negative $\Delta G^0$,it must have the most positive $E^0_{cell}$ value.
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
$A) \ E^0_{cell} = 0.80 - (-0.76) = 1.56 \ V; \Delta G^0 = -2 \times F \times 1.56 = -3.12 \ F$
$B) \ E^0_{cell} = -2.37 - (-0.76) = -1.61 \ V; \Delta G^0 = -2 \times F \times (-1.61) = +3.22 \ F$
$C) \ E^0_{cell} = -2.37 - 0.80 = -3.17 \ V; \Delta G^0 = -2 \times F \times (-3.17) = +6.34 \ F$
$D) \ E^0_{cell} = 0.80 - 0.34 = 0.46 \ V; \Delta G^0 = -2 \times F \times 0.46 = -0.92 \ F$
Comparing the values,option $(A)$ gives the most negative $\Delta G^0$.