Write the anode, cathode and overall reaction involved in a dry cell.
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Solution and Explanation
Step 1: At the anode (zinc loses electrons)
The zinc gives up electrons and goes into solution as $\mathrm{Zn^{2+}}$:\[ Zn \rightarrow Zn^{2+} + 2e^- \]
Step 2: At the cathode (manganese is reduced)
The electrons travel to the cathode where $\mathrm{MnO_2}$ is reduced with the help of the ammonium ions:\[ 2MnO_2 + 2NH_4^+ + 2e^- \rightarrow Mn_2O_3 + 2NH_3 + H_2O \]
Step 3: Add them for the overall reaction
\[ Zn + 2MnO_2 + 2NH_4^+ \rightarrow Zn^{2+} + Mn_2O_3 + 2NH_3 + H_2O \]
Answer: Anode: $Zn \rightarrow Zn^{2+} + 2e^-$; Cathode: $2MnO_2 + 2NH_4^+ + 2e^- \rightarrow Mn_2O_3 + 2NH_3 + H_2O$; Overall: $Zn + 2MnO_2 + 2NH_4^+ \rightarrow Zn^{2+} + Mn_2O_3 + 2NH_3 + H_2O$.
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