Calculate the standard cell potentials of galvanic cell in which the following reactions take place (i) $2 \mathrm{Cr}(s)+3 \mathrm{Cd}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Cd}(s)$ (ii) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ag}(s)$ $\mathrm{E}_{\mathrm{Cd}^{2+}, \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V}$, $\mathrm{E}_{\mathrm{Ag}^{+}, \mathrm{Ag}}^{\circ}=\mathbf{0 . 8 0 \mathrm { V } ,} \mathrm{E}_{\mathrm{Fe}^{3+}}^{\circ}, \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$. Calculate the $\Delta_r G^{\circ}$ and equilibrium constant of the reactions.

Question

Calculate the standard cell potentials of galvanic cell in which the following reactions take place (i) $2 \mathrm{Cr}(s)+3 \mathrm{Cd}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Cd}(s)$ (ii) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ag}(s)$ $\mathrm{E}_{\mathrm{Cd}^{2+}, \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V}$, $\mathrm{E}_{\mathrm{Ag}^{+}, \mathrm{Ag}}^{\circ}=\mathbf{0 . 8 0 \mathrm { V } ,} \mathrm{E}_{\mathrm{Fe}^{3+}}^{\circ}, \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$. Calculate the $\Delta_r G^{\circ}$ and equilibrium constant of the reactions.,

MARKS App · Accepted Answer

(i) $\quad \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}$ $\begin{aligned} &=-0.40 \mathrm{~V}-(-0.74 \mathrm{~V})=+0.34 \mathrm{~V} \ &\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ} \ &=-6 \times 96500 \mathrm{C} \mathrm{mol}^{-1} \times 0.34 \mathrm{~V} \ &=-196860 \mathrm{C} \mathrm{V} \mathrm{mol}^{-1} \ &=-196860 \mathrm{~J} \mathrm{~mol}^{-1}=-196.86 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \ &-\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=2.303 \mathrm{RT} \operatorname{log~K} \ &196860=2.303 \times 8.314 \times 298 \operatorname{log~K} \ &\text { or } \quad \log \mathrm{K}=34.5014 \ &\mathrm{~K}=\text { Antilog } 34.5014=3.172 \times 10^{34} \end{aligned}$ (ii) $\mathrm{E}_{\text {cell }}^0=+0.80 \mathrm{~V}-0.77 \mathrm{~V}=+0.03 \mathrm{~V}$ $\begin{aligned} &\Delta_r G^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ} \ &=-1 \times\left(96500 \mathrm{C} \mathrm{mol}^{-1}\right) \times(0.03 \mathrm{~V}) \ &=-2895 \mathrm{C} \mathrm{V} \mathrm{mol}^{-1}=-2895 \mathrm{~J} \mathrm{~mol}^{-1} \ &=-2.895 \mathrm{~kJ} \mathrm{~mol}^{-1} \ &\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \operatorname{log~K} \ &-2895=-2.303 \times 8.314 \times 298 \times \operatorname{log~K} \ &\text { or } \operatorname{log~K}=0.5074 \ &\text { or } \mathrm{K}=\text { Antilog }(0.5074)=3.22 . \end{aligned}$

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