Consider the following $\mathrm{E}^{\circ}$ values $$ \begin{aligned} & \mathrm{E}_{\mathrm{Fe}^{3+} /…

Consider the following $\mathrm{E}^{\circ}$ values $$ \begin{aligned} & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ Under standard conditions the potential for the reaction $\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is
  1. $1.68 \mathrm{~V}$
  2. $0.63 \mathrm{~V}$
  3. $0.91 \mathrm{~V}$
  4. $1.40 \mathrm{~V}$

Solution

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\mathrm{RHS}}^{\circ}-\mathrm{E}_{\mathrm{LHS}}^{\circ}$ $=(0.77)-(-0.14)$ $=0.91 \mathrm{~V}$

Asked in: JEE Main 2004

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