NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Hard

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Analyzing +4 | -1

What is

E^{\mathrm{o}}

for the following half reaction?

\begin{aligned} & \mathrm{MX}_2(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s})+2 \mathrm{X}^{-} ; E^{\circ}=? \\ & \mathrm{M}^{2+}(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s}) ; E^\circ=0.100 \mathrm{~V} \\ & \mathrm{MX}_2(\mathrm{s}) \rightarrow \mathrm{M}^{2+}+2 \mathrm{X}^{-} ; K_{\mathrm{sp}}=e^{-10} \end{aligned}

View answer and explanation

Correct option: B

E^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}

Option B (

E^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}

) is correct because it correctly applies the Nernst equation to relate the standard potential to the solubility product.

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