Question

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}

Write a balanced equation for the cell’s reaction.

Solution

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Answered 2 years ago
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Firstly, we should determine the reduction and oxidation half cell reaction in this battery. The half cell with the lower reduction potential will proceed in the opposite direction. In this case, tin is where oxidation occurs with the following half-cell reaction:

SnSn2++2e\begin{align*} Sn \rightarrow Sn^{2+} +2e^{-} \end{align*}

The reduction will occur in the half cell with the higher reduction potential which is the mercury (II) ion with the following half-cell reactions:

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

Since the electrons are balanced in the two half-cell reactions. Add the equation and cancel out electrons.

SnSn2++2eHg2++2eHgSn+Hg2+Sn2++Hg\begin{align*} Sn \rightarrow Sn^{2+} +2e^{-} \\ Hg^{2+} + 2e^{-} \rightarrow Hg \\ Sn + Hg^{2+} \rightarrow Sn^{2+} + Hg \end{align*}

Thus, our balanced overall cell reaction is:

Sn+Hg2+Sn2++Hg\begin{align*} Sn + Hg^{2+} \rightarrow Sn^{2+} + Hg \end{align*}

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