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A battery is assembled using tin and mercury, which have the following reduction half-reactions:

Sn2++2eSnHg2++2eHg\begin{array}{l} \mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn} \\ \mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg} \end{array}

Which reaction occurs at the anode? At the cathode?

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The reaction that occurs at the anode is at tin eletrode where the oxidation reaction takes place. SnSn is oxidized by losing 2 electrons from SnSn to Sn2+Sn^{2+}. This half-cell also has a lower reduction potential and is more negative with the oxidation half cell reaction of:

 Sn2e+Sn2+\begin{align*} Sn \rightarrow 2e^{-} + Sn^{2+} \end{align*}

The mercury electrode is the cathode, Hg2+Hg^{2+} is reduced from Hg2+Hg^{2+} to HgHg by gaining electrons. This is where reduction reaction takes place. This electrode has higher reduction potential and more positive with the following half cell reaction:

 Hg2++2eHg\begin{align*} Hg^{2+} +2e^{-} \rightarrow Hg\end{align*}

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