For the given reactions
Sn2+ + 2e$-$ $\to$ Sn
Sn4+ + 4e$-$ $\to$ Sn
the electrode potentials are ; $E_{S{n^{2 + }}/Sn}^o = - 0.140$ V and $E_{S{n^{4 + }}/Sn}^o = + 0.010$ V. The magnitude of standard electrode potential for $S{n^{4 + }}/S{n^{2 + }}$ i.e. $E_{S{n^{4 + }}/S{n^{2 + }}}^o$ is _____________ $\times$ 10$-$2 V. (Nearest integer)
Answer (integer)
16
Solution
$\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \quad \mathrm{E}_{1}^{0}=0.140 \mathrm{~V}$
<br/><br/>
$$
\mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V}
$$
<br/><br/>
$$
\begin{aligned}
& \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\
& \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\
& \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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