Adding powdered \( \mathrm{Pb} \) and \( \mathrm{Fe} \) to a solution containing \( 1.0 \mathrm{...

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Adding powdered \( \mathrm{Pb} \) and \( \mathrm{Fe} \) to a solution containing \( 1.0 \mathrm{M} \) is each of \( \mathrm{Pb}^{2+} \) and \( \mathrm{Fe}^{2+} \) ions would result into the formation of Given,
\( \begin{array}{l} \mathrm{E}_{\mathrm{Pb}^{+2} / \mathrm{Pb}}^{\circ}=-0.13 \mathrm{~V} \\ \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V} \end{array} \)
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