$Z n | Z n^{2 +} (C_{1}) | | Z n^{2 +} (C_{2}) | Z n$ . For this cell $Δ G$ is negative if:

C_{1} = C_{2}

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C_{1} > C_{2}

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C_{2} > C_{1}

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Solution

The correct option is C $C_{2} > C_{1}$
$Z n \to Z n^{2 +} (C_{1}) + 2 e^{-}$
$Z n^{2 +} (C_{2}) + 2 e^{-} \to Z n$
$Z n^{2 +} (C_{2}) ⇌ Z n^{2 +} (C_{1})$
$E = E_{cell}^{\circ} - \frac{0.0591}{2} log \frac{C_{1}}{C_{2}}$
$E = 0 - \frac{0.0591}{2} log \frac{C_{1}}{C_{2}}$
For $E \to +$ ve when $C_{1} < C_{2}$
As we know $Δ G = - n F E_{cell}$
$E \to +$ ve then $Δ G \to -$ ve

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