Mix Examples-Electrochemistry Questions in English

(A) metal can liberate $H_2$ gas from an acid if its standard reduction potential is lower than that of $H^{+}/H_2$ $(0.0 \ V)$.
$I$. $Zn$ $(E^{\circ} = -0.76 \ V)$ reacts with $HCl$ to liberate $H_2$ gas because $-0.76 \ V < 0.0 \ V$.
$II$. $Cu$ $(E^{\circ} = 0.34 \ V)$ does not react with $HCl$ because its reduction potential is higher than $0.0 \ V$.
$III$. $Cu$ reacts with $HNO_3$ to produce $NO$ gas instead of $H_2$ gas because $NO_3^{-}$ is a stronger oxidizing agent than $H^{+}$.
Therefore,reactions $II$ and $III$ do not liberate $H_{2(g)}$.

(B) substance with a lower (more negative) standard reduction potential acts as a stronger reducing agent. Conversely,a substance with a higher (more positive) standard reduction potential acts as a stronger oxidizing agent.

$Zn^{2+}/Zn$	$-0.76 \ V$
$2H^{+}/H_2$	$0.00 \ V$
$Cu^{2+}/Cu$	$0.34 \ V$
$NO_3^{-}, H^{+}/NO$	$0.97 \ V$

$I.$ Since $E^{\circ}_{Cu^{2+}/Cu} (0.34 \ V) > E^{\circ}_{2H^{+}/H_2} (0.0 \ V)$,$H^{+}$ cannot oxidize $Cu$ to $Cu^{2+}$. Statement $I$ is correct.
$II.$ Since $E^{\circ}_{Zn^{2+}/Zn} (-0.76 \ V) < E^{\circ}_{Cu^{2+}/Cu} (0.34 \ V)$,$Zn$ acts as a stronger reducing agent and can reduce $Cu^{2+}$ to $Cu$. Statement $II$ is correct.
$III.$ Since $E^{\circ}_{NO_3^{-}, H^{+}/NO} (0.97 \ V) > E^{\circ}_{Cu^{2+}/Cu} (0.34 \ V)$,$NO_3^{-}$ can oxidize $Cu$ to $Cu^{2+}$. Statement $III$ is correct.
Therefore,all statements $I, II,$ and $III$ are correct.

(B) Given: $n = 2$,$T = 300 \ K$,$E^{\circ} = 1.0 \ V$,$\Delta_{r}H^{\circ} = -163 \ kJ \ mol^{-1}$.
Using the relation $\Delta G^{\circ} = -nFE^{\circ}$:
$\Delta G^{\circ} = -2 \times 96500 \times 1.0 \ J \ mol^{-1} = -193000 \ J \ mol^{-1} = -193 \ kJ \ mol^{-1}$.
We know that $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Rearranging for $\Delta S^{\circ}$: $\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}$.
Substituting the values: $\Delta S^{\circ} = \frac{-163 - (-193)}{300} \ kJ \ K^{-1} \ mol^{-1} = \frac{30}{300} \ kJ \ K^{-1} \ mol^{-1} = 0.1 \ kJ \ K^{-1} \ mol^{-1}$.
Converting to $J \ K^{-1} \ mol^{-1}$: $\Delta S^{\circ} = 0.1 \times 1000 \ J \ K^{-1} \ mol^{-1} = 100 \ J \ K^{-1} \ mol^{-1}$.

(A) Given: $\Delta_r S^{\circ} = 100 \ J \ K^{-1} \ mol^{-1}$,$E_{cell}^{\circ} = 1.0 \ V$,$n = 2$,$T = 300 \ K$,$F = 96500 \ C \ mol^{-1}$.
Using the relation $\Delta G^{\circ} = -nFE_{cell}^{\circ}$:
$\Delta G^{\circ} = -2 \times 96500 \times 1 = -193000 \ J \ mol^{-1} = -193 \ kJ \ mol^{-1}$.
Using the Gibbs-Helmholtz equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
$\Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ}$.
$\Delta H^{\circ} = -193 \ kJ \ mol^{-1} + (300 \ K \times 100 \ J \ K^{-1} \ mol^{-1}) / 1000$.
$\Delta H^{\circ} = -193 \ kJ \ mol^{-1} + 30 \ kJ \ mol^{-1} = -163 \ kJ \ mol^{-1}$.

(NONE) The electrochemical equivalent $(Z)$ is defined by the relation $w = Z \cdot I \cdot t$,where $w$ is the mass in grams,$I$ is current in Amperes,and $t$ is time in seconds.
Since $1 \ Coulomb = 1 \ Ampere \cdot 1 \ second$,the unit of $Z$ is $g/Coulomb$.
All statements $A$,$B$,$C$,and $D$ are scientifically correct.
However,if this is a multiple-choice question where one must be incorrect,there might be a context-specific error. Given the standard definitions:
$A$ is correct as $NaCl$ dissociates into ions.
$B$ is correct as $Z = w / (I \cdot t)$.
$C$ is correct as $n$ is the stoichiometric coefficient of electrons.
$D$ is correct by $IUPAC$ convention.
Since all are correct,the question is technically flawed.

(A) The correct matching is as follows:
$A$. Leclanche cell: The anode reaction is $Zn \longrightarrow Zn^{2+} + 2e^{-}$. Thus,$A-3$.
$B$. Fuel cell: It converts the energy of combustion of fuels like $H_2$ directly into electrical energy. Thus,$B-1$.
$C$. Ni-Cd cell: It is a rechargeable (secondary) cell. Thus,$C-2$.
Therefore,the correct sequence is $A-3, B-1, C-2$.

(D) First,calculate the limiting molar conductivity of methanoic acid $(HCOOH)$:
$\Lambda_{m}^{\circ}(HCOOH) = \lambda_{H^{+}}^0 + \lambda_{HCOO^{-}}^0 = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^0} = \frac{40.42}{404.2} = 0.1$
Finally,calculate the dissociation constant $(K_{a})$:
$K_{a} = \frac{C\alpha^2}{1-\alpha} = \frac{0.027 \times (0.1)^2}{1-0.1} = \frac{0.027 \times 0.01}{0.9} = \frac{0.00027}{0.9} = 3.0 \times 10^{-4}$

(B) Given: Cell constant $(G^*)$ = $0.5 \ cm^{-1}$,Conductance $(G)$ = $3.12 \times 10^{-4} \ S$,Concentration $(C)$ = $0.01 \ M$.
$\text{Conductivity } (\kappa) = G \times G^* = 3.12 \times 10^{-4} \ S \times 0.5 \ cm^{-1} = 1.56 \times 10^{-4} \ S \ cm^{-1}$.
$\text{Molar conductivity } (\Lambda_m) = \frac{\kappa \times 1000}{C} = \frac{1.56 \times 10^{-4} \times 1000}{0.01} = 15.6 \ S \ cm^2 \ mol^{-1}$.
$\text{Limiting molar conductivity } (\Lambda_m^\circ) = \lambda_m^\circ(H^+) + \lambda_m^\circ(CH_3COO^-) = 349 + 41 = 390 \ S \ cm^2 \ mol^{-1}$.
$\text{Degree of dissociation } (\alpha) = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{15.6}{390} = 0.04$.
$\text{Dissociation constant } (K_a) = \frac{C \alpha^2}{1 - \alpha} = \frac{0.01 \times (0.04)^2}{1 - 0.04} = \frac{0.01 \times 0.0016}{0.96} = \frac{1.6 \times 10^{-5}}{0.96} \approx 1.67 \times 10^{-5}$.

(A) The molar conductivity at infinite dilution for acetic acid is given by Kohlrausch's law: $\Lambda_m^0(CH_3COOH) = \lambda_m^0(H^+) + \lambda_m^0(CH_3COO^-) = 349 + 41 = 390 \ S \ cm^2 \ mol^{-1}$.
Degree of dissociation $\alpha = \frac{\Lambda_m^c}{\Lambda_m^0} = \frac{7.8}{390} = 0.02$.
The dissociation constant $K_a$ is given by $K_a = \frac{c \alpha^2}{1 - \alpha}$.
Given $c = 0.04 \ M$,$K_a = \frac{0.04 \times (0.02)^2}{1 - 0.02} = \frac{0.04 \times 0.0004}{0.98} = \frac{0.000016}{0.98} \approx 1.63 \times 10^{-5}$.

(C) Moles of $CH_3COOH = 0.2 \text{ M} \times 0.1 \text{ L} = 0.02 \text{ mol}$.
Since $CH_3COOH$ is completely neutralized by $NaOH$, moles of $CH_3COONa$ formed $= 0.02 \text{ mol}$.
According to Kolbe's electrolysis reaction:
$2CH_3COONa + 2H_2O \rightarrow CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$.
From the stoichiometry, $2 \text{ moles}$ of $CH_3COONa$ produce $1 \text{ mole}$ of ethane $(C_2H_6)$.
Moles of $C_2H_6 = \frac{0.02}{2} = 0.01 \text{ mol}$.
Volume of $C_2H_6$ at $STP = 0.01 \text{ mol} \times 22.4 \text{ L/mol} = 0.224 \text{ L}$.

(C) The standard Gibb's free energy change $(\Delta G^{\circ})$ is calculated using the formula: $\Delta G^{\circ} = -RT \ln K_C = -2.303 RT \log K_C$.
Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$,$T = 25 + 273 = 298 \ K$,and $K_C = 10^{12}$.
Substituting the values: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log(10^{12})$.
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times 12$.
$\Delta G^{\circ} = -68,470.18 \ J \ mol^{-1} = -68.47 \ kJ \ mol^{-1}$.

(B) The degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.
Given $K_a = 1.8 \times 10^{-5}$ and $C = 0.01 \ M = 10^{-2} \ M$.
$\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{10^{-2}}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{18 \times 10^{-4}} = 4.24 \times 10^{-2} = 0.0424$.
The molar conductivity $\Lambda_m$ is related to the molar conductivity at infinite dilution $\Lambda_m^\circ$ by $\Lambda_m = \alpha \times \Lambda_m^\circ$.
$\Lambda_m = 0.0424 \times 390 \ S \ cm^2 \ mol^{-1} = 16.536 \ S \ cm^2 \ mol^{-1} \approx 16.54 \ S \ cm^2 \ mol^{-1}$.

(C) For a hydrogen electrode,$E = -0.0591 \times pH = -0.0591 \times 10 = -0.591 \ V$. Thus,statement $(A)$ is correct.
$(B)$ $\Lambda^{\circ}_{m}(CaCl_2) = \Lambda^{\circ}_{m}(Ca^{2+}) + 2 \times \Lambda^{\circ}_{m}(Cl^{-}) = 119 + 2(76) = 119 + 152 = 271 \ S \ cm^2 \ mol^{-1}$. Since $271 \neq 195$,statement $(B)$ is incorrect.
$(C)$ The Nernst equation at equilibrium is $E_{cell} = 0$,which gives $0 = E_{cell}^{0} - \frac{2.303 RT}{nF} \log K_{c}$,so $E_{cell}^{0} = \frac{2.303 RT}{nF} \log K_{c}$. Thus,statement $(C)$ is correct.
Therefore,statements $(A)$ and $(C)$ are correct.

(B) The cell reaction is: $A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$
Here,$A$ is oxidized to $A^{+}$ $(n=1)$ and $B^{+}$ is reduced to $B$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{B^{+}/B} - E^{\circ}_{A^{+}/A} = 0.5 - (-0.5) = 1.0 \ V$
Using the relation: $E^{\circ}_{cell} = \frac{0.06}{n} \log K_C$
Substituting the values: $1.0 = \frac{0.06}{1} \log K_C$
Therefore,$\log K_C = \frac{1.0}{0.06} = \frac{100}{6}$

(B) Given the Latimer diagram:
$BrO_3^{-}$ $\xrightarrow{E_1^{\circ}} BrO^{-}$ $\xrightarrow{0.45 \ V} \frac{1}{2} Br_2$ $\xrightarrow{1.07 \ V} Br^{-}$
And the overall potential $BrO_3^{-} \rightarrow Br^{-}$ is $0.61 \ V$.
We use the relation $\Delta G^{\circ} = -nFE^{\circ}$.
For the overall reaction $BrO_3^{-} \rightarrow Br^{-}$:
$n = 6$,so $\Delta G_{total}^{\circ} = -6F(0.61 \ V) = -3.66F$.
For the individual steps:
$1$. $BrO_3^{-} \rightarrow BrO^{-}$: $n = 4$,$\Delta G_1^{\circ} = -4FE_1^{\circ}$.
$2$. $BrO^{-} \rightarrow \frac{1}{2} Br_2$: $n = 1$,$\Delta G_2^{\circ} = -1F(0.45 \ V) = -0.45F$.
$3$. $\frac{1}{2} Br_2 \rightarrow Br^{-}$: $n = 1$,$\Delta G_3^{\circ} = -1F(1.07 \ V) = -1.07F$.
Summing the $\Delta G^{\circ}$ values:
$-3.66F = -4FE_1^{\circ} - 0.45F - 1.07F$
$-3.66 = -4E_1^{\circ} - 1.52$
$4E_1^{\circ} = 3.66 - 1.52$
$4E_1^{\circ} = 2.14$
$E_1^{\circ} = \frac{2.14}{4} = 0.535 \ V$.

(C) For hydrogen electrode: $E = E^0 - 0.0591 \ pH = 0 - 0.0591 \times 10 = -0.591 \ V$. Thus,$A-III$.
$(B)$ Standard reduction potential of $Cu^{2+}|Cu$ is $0.337 \ V$. Thus,$B-IV$.
$(C)$ Standard oxidation potential of $Zn|Zn^{2+}$ is $0.76 \ V$. Thus,$C-I$.
$(D)$ The value of $\frac{2.303RT}{F}$ at $298 \ K$ is $0.059$. Thus,$D-II$.
The correct matching is $A-III, B-IV, C-I, D-II$,which corresponds to option $(c)$.

(B) The unit of electrochemical equivalent $(Z)$ is $g/Coulomb$.
From Faraday's law,$w = Z \cdot I \cdot t$.
Therefore,$Z = \frac{w}{I \cdot t}$,which has units of $g/Coulomb$.
Thus,the statement in option $B$ is incorrect as it states the units are $g \cdot Coulomb$ instead of $g/Coulomb$.

(B) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$
First,calculate the molar conductivity $(\Lambda_m)$ using the formula: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given: $\kappa = 19.5 \times 10^{-5} \ S \ cm^{-1}$ and $C = 0.01 \ mol \ L^{-1}$
$\Lambda_m = \frac{19.5 \times 10^{-5} \times 1000}{0.01} = 19.5 \ S \ cm^2 \ mol^{-1}$
Now,calculate $\alpha$:
$\alpha = \frac{19.5}{390} = 0.05 = 5.0 \times 10^{-2}$

(A) $1$. The molar conductivity of $BaSO_4$ at infinite dilution is $\Lambda_m^\circ (BaSO_4) = \Lambda_m^\circ(Ba^{2+}) + \Lambda_m^\circ(SO_4^{2-})$.
Using Kohlrausch's law:
$\Lambda_m^\circ(BaSO_4) = \Lambda_m^\circ(BaCl_2) + \Lambda_m^\circ(H_2SO_4) - 2\Lambda_m^\circ(HCl) = x_1 + x_2 - 2x_3$.
$2$. The solubility $S$ (in $\text{mol L}^{-1}$) is given by $S = \frac{1000 \cdot x}{\Lambda_m^\circ} = \frac{1000 \cdot x}{x_1 + x_2 - 2x_3}$.
$3$. The solubility product $K_{sp}$ for $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$ is $K_{sp} = S^2$.
Substituting the value of $S$:
$K_{sp} = \left(\frac{1000 x}{x_1 + x_2 - 2x_3}\right)^2 = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$.

(A) The overall cell reaction is obtained by subtracting the first half-reaction from the second (multiplied by $3$):
$3 \times (\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O) \implies 1.5O_2 + 6H^+ + 6e^- \rightarrow 3H_2O$ $(E^{\ominus} = 1.23 \text{ V})$
$CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^-$ $(E^{\ominus} = -0.02 \text{ V})$
Adding these,we get: $CH_3OH + 1.5O_2 \rightarrow CO_2 + 2H_2O$.
$E^{\ominus}_{cell} = 1.23 \text{ V} - 0.02 \text{ V} = 1.21 \text{ V}$.
The number of electrons transferred,$n = 6$.
The standard Gibbs free energy change is $\Delta G^{\ominus} = -nFE^{\ominus} = -6 \times 96500 \text{ C mol}^{-1} \times 1.21 \text{ V} = -700770 \text{ J mol}^{-1}$.
The work derived from the cell at $80\%$ efficiency is $W = 0.8 \times |\Delta G^{\ominus}| = 0.8 \times 700770 \text{ J} = 560616 \text{ J}$.
Work done in isothermal compression against constant pressure is $W = P \Delta V$.
Given $P = 1 \text{ kPa} = 1000 \text{ Pa}$,we have $560616 = 1000 \times \Delta V$.
$\Delta V = 560.616 \text{ m}^3$.
Rounding to the nearest integer,$\Delta V = 561 \text{ m}^3$.