The standard free energy change $\left(\Delta G^{\circ}\right)$ for the following reaction (in kJ ) at…

The standard free energy change $\left(\Delta G^{\circ}\right)$ for the following reaction (in kJ ) at $25^{\circ} \mathrm{C}$ is $3 \mathrm{Ca}(\mathrm{s})+2 \mathrm{Au}^{3+}$ (aq. 1 M ) $\rightarrow$ $3 \mathrm{Ca}^{2+}(\mathrm{aq}, 1 \mathrm{M})+2 \mathrm{Au}(\mathrm{s})$ (given : $\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\mathrm{o}}=+1.50 \mathrm{~V}, \mathrm{E}_{\mathrm{Ca}^{2+} / \mathrm{Ca}}^0=-2.87 \mathrm{~V}$, $1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}$ )
  1. $-2.53 \times 10^3$
  2. $+2.53 \times 10^3$
  3. $-2.53 \times 10^4$
  4. $+2.53 \times 10^4$

Solution

$\begin{aligned} & \text { } 3 \mathrm{Ca}(\mathrm{~s})+2 \mathrm{Au}^{3+}(\mathrm{aq} 1 \mathrm{M}) \rightarrow 3 \mathrm{Ca}^{2+}(\mathrm{aq} \mathrm{1M})+2 \mathrm{Au}(\mathrm{~S}) \\ & \mathrm{E}^0=\mathrm{E}_{\mathrm{L}}+\mathrm{E}_{\mathrm{R}} \\ & =\mathrm{E}_{\mathrm{Ca} / \mathrm{Ca}^{2+}}+\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}} \\ & =2.87+1.50 \\ & \mathrm{E}^{\mathrm{o}}=4.37 \mathrm{~V} \end{aligned}$
Total no. of electron envolved by balancing the electron. $\begin{aligned} & \mathrm{n}=6 \\ & \therefore \quad \Delta \mathrm{G}^{\mathrm{o}}=-\mathrm{nFE}^{\mathrm{o}} \\ & =-6 \times 96500 \times 4.37 \mathrm{~mol} \cdot \mathrm{C} \mathrm{~mol}^{-1} \cdot \mathrm{~V} \\ & =-2530230 \mathrm{CV} \quad[1 \mathrm{~J}=1 \mathrm{CV}] \\ & =-2530.23 \mathrm{~kJ} \quad\left[1 \mathrm{~kJ}=10^3 \mathrm{~J}\right] \\ & \Delta \mathrm{G}^{\mathrm{o}}=-2.53 \times 10^3 \mathrm{~kJ} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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