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The standard potential at $25^{0} C$ for the following half reactions is given :
$\begin{array}{l} Z n^{+ 2} + 2 e^{-} \to Z n; E^{0} = - 0.762 V \\ M g^{2 +} + 2 e^{-} \to M g; E^{0} = - 2.37 V \end{array}$
When zinc dust is added to the solutions of $M g C l_{2}$ .

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$Z n C l_{2}$ is formed

$M g$ is precipitated

$Z n$ dissolved in the solution

No reactions takes place.

answer is D.

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Detailed Solution

${E^{0}}_{Z n | Z n^{+ 2}} + {E^{0}}_{M g^{2 +} | M g} = - 1.608$

( Non-spontaneous

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