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NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Hard
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Analyzing +4 | -1
Consider the reaction of extraction of gold from its ore:

Au(s)+2CN−(aq)+14O2(g)+12H2O(l)→Au(CN)2−(aq)+OH−(aq)\begin{aligned} & \mathrm{Au}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq})+\frac{1}{4} \mathrm{O}_2(\mathrm{g})+\frac{1}{2} \mathrm{H}_2 \mathrm{O}({l}) \rightarrow \mathrm{Au}(\mathrm{CN})_2{ }^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}

Use the following data to calculate ΔG∘\Delta G^{\circ} for the above reaction.

Kf[Au(CN)2−]=XO2+2H2O+4e−→4OH−;E∘=+0.41 VAu3++3e−→Au;E∘=+1.50 VAu3++2e−→Au+;E∘=+1.40 V\begin{aligned} & K_{\mathrm{f}}\left[\mathrm{Au}(\mathrm{CN})_2^{-}\right] =X \\ & \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-} ; E^{\circ}=+0.41 \mathrm{~V} \\ & \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; E^{\circ}=+1.50 \mathrm{~V} \\ & \mathrm{Au}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Au}^{+} ; E^{\circ}=+1.40 \mathrm{~V} \end{aligned}
View answer and explanation
Correct option: A
−RTln⁡X+1.29 F-R T \ln X+1.29 \mathrm{~F}
Option A is correct because the overall reaction involves the formation of a complex ion and a redox reaction, and the Gibbs free energy change can be calculated using the given data.

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