Calculate the equilibrium constant at $298 \ K$ for the following reaction and also calculate the maximum work that can be obtained from this cell:
$Mg(s) \ | \ Mg^{2+}(aq) \ || \ Ag^{+}(aq) \ | \ Ag(s)$
Given: $E_{Mg^{2+} \mid Mg}^{o} = -2.37 \ V$ and $E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V$
$Mg(s) \ | \ Mg^{2+}(aq) \ || \ Ag^{+}(aq) \ | \ Ag(s)$
Given: $E_{Mg^{2+} \mid Mg}^{o} = -2.37 \ V$ and $E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V$
$1$. Calculation of $E_{\text{cell}}^{o}$:
$E_{\text{cell}}^{o} = E_{\text{cathode}}^{o} - E_{\text{anode}}^{o} = E_{Ag^{+} \mid Ag}^{o} - E_{Mg^{2+} \mid Mg}^{o}$
$E_{\text{cell}}^{o} = 0.80 \ V - (-2.37 \ V) = 3.17 \ V$
$2$. Calculation of equilibrium constant $K_{C}$:
Using the formula $E_{\text{cell}}^{o} = \frac{0.0591}{n} \log K_{C}$,where $n = 2$:
$3.17 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{3.17 \times 2}{0.0591} \approx 107.2758$
$K_{C} = \text{antilog}(107.2758) \approx 1.887 \times 10^{107}$
$3$. Calculation of maximum work $(W_{\text{max}})$:
Maximum work is equal to the decrease in Gibbs free energy,$W_{\text{max}} = -\Delta G^{o} = nFE_{\text{cell}}^{o}$
$W_{\text{max}} = 2 \times 96487 \ C \ mol^{-1} \times 3.17 \ V$
$W_{\text{max}} = 611727.6 \ J \ mol^{-1} = 611.73 \ kJ \ mol^{-1}$
$E_{\text{cell}}^{o} = E_{\text{cathode}}^{o} - E_{\text{anode}}^{o} = E_{Ag^{+} \mid Ag}^{o} - E_{Mg^{2+} \mid Mg}^{o}$
$E_{\text{cell}}^{o} = 0.80 \ V - (-2.37 \ V) = 3.17 \ V$
$2$. Calculation of equilibrium constant $K_{C}$:
Using the formula $E_{\text{cell}}^{o} = \frac{0.0591}{n} \log K_{C}$,where $n = 2$:
$3.17 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{3.17 \times 2}{0.0591} \approx 107.2758$
$K_{C} = \text{antilog}(107.2758) \approx 1.887 \times 10^{107}$
$3$. Calculation of maximum work $(W_{\text{max}})$:
Maximum work is equal to the decrease in Gibbs free energy,$W_{\text{max}} = -\Delta G^{o} = nFE_{\text{cell}}^{o}$
$W_{\text{max}} = 2 \times 96487 \ C \ mol^{-1} \times 3.17 \ V$
$W_{\text{max}} = 611727.6 \ J \ mol^{-1} = 611.73 \ kJ \ mol^{-1}$
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$B.$ $E_{cell}^0$ is dependent on the pressure.
$C.$ $\frac{dE_{cell}^0}{dT} = \frac{\Delta_{r}S^0}{nF}$.
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$A.$ The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.
$B.$ $E_{cell}^0$ is dependent on the pressure.
$C.$ $\frac{dE_{cell}^0}{dT} = \frac{\Delta_{r}S^0}{nF}$.
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View SolutionAn aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)
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View SolutionMatch the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+}|Cu$ $(II)$ $0.059$ $(C)$ $Zn|Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+}|Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn|Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
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