Consider the cell reaction, at 300 K $\mathrm{A}(s)+\mathrm{B}^{2+}(a q) \rightleftharpoons…

Consider the cell reaction, at 300 K $\mathrm{A}(s)+\mathrm{B}^{2+}(a q) \rightleftharpoons \mathrm{A}^{2+}(a q)+\mathrm{B}(s)$ Its $\mathrm{E}^{\circ}$ is 1.0 V . The $\Delta_{\mathrm{r}} \mathrm{H}^{\circ}$ of the reaction is $-163 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What is $\Delta_{\mathrm{r}} \mathrm{S}^{\circ}$ (in J K${ }^{-1}$ ) of the reaction? $\left(\mathrm{F}=96500 \mathrm{C} \mathrm{~mol}^{-1}\right)$
  1. 10
  2. 100
  3. 1000
  4. 10000

Solution

Given, $\begin{aligned} & \mathrm{E}^{\circ}=1.0 \mathrm{~V} \\ & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-163 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{n}=2 \\ & \mathrm{~T}=300 \mathrm{k} \\ & \therefore \quad \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}{ }^{\circ} \\ & =-2 \times 96500 \times 1.0 \\ & =193000 \mathrm{CV} \quad[1 \mathrm{~J}=1 \mathrm{CV}] \\ & =\frac{193000}{1000} \mathrm{~kJ} \quad\left[1 \mathrm{~kJ}=10^3 \mathrm{~J}\right] \\ & \Delta \mathrm{G}^{\circ}=193 \mathrm{~kJ} \\ & \therefore \quad \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S} \\ & \text { or, } \Delta \mathrm{S}=\frac{\Delta \mathrm{H}-\Delta \mathrm{G}}{\mathrm{~T}} \\ & \Rightarrow \Delta S=\frac{-163+193}{300}=\frac{30}{300}=0.1 \mathrm{~kJ} \mathrm{~K}^{-1} \\ & \therefore \quad \Delta \mathrm{~S}=0.1 \times 10^3 \mathrm{JK}^{-1} \\ & \Delta \mathrm{~S}=100 \mathrm{JK}^{-1} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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