Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of $E_{\text {red }}^{\circ}\( for the cathode reaction is \)+0.098 \mathrm{~V}$. The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

To determine the overall cell reaction, we need to combine the two half-cell reactions. In the battery, the two half-cell reactions are \[ \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \] and \[ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}. \] We need to add these two equations to cancel out the electrons while keeping all the other species involved in the reactions. Let's add them up: \[ \begin{array}{ l l } \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} &\longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ + & \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) &\longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2}\mathrm{O}(l)+2 \mathrm{e}^{-} \\ \hline \mathrm{HgO}(s) + \mathrm{Zn}(s)+ &2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s)+ \\ \mathrm{H}_{2}\mathrm{O}(l) + & \\ \end{array}. \] Now the overall cell reaction is \[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s). \]

The overall cell potential is the difference between the standard reduction potentials for the two half-cell reactions: \[ E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}. \] The standard reduction potential for the cathode reaction, \(E_{\mathrm{cathode}}^{\circ}\), is given as \(+0.098 \mathrm{~V}\). The overall cell potential, \(E_{\mathrm{cell}}^{\circ}\), is given as \(+1.35 \mathrm{~V}\). We can solve for the standard reduction potential for the anode reaction, \(E_{\mathrm{anode}}^{\circ}\): \[ E_{\mathrm{anode}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{cell}}^{\circ} \] \[ E_{\mathrm{anode}}^{\circ} = +0.098 \mathrm{~V} - (+1.35 \mathrm{~V}) \] \[ E_{\mathrm{anode}}^{\circ} = - 1.252 \mathrm{~V} \] The standard reduction potential for the anode reaction is \(-1.252 \mathrm{~V}\).

In this exercise, we see that the anode reaction does not occur in an acidic medium but in a basic medium. The reaction involves hydroxide ions (\(\mathrm{OH}^{-}\)), which are present in basic solutions. In an acidic medium, the reaction would involve hydrogen ions (\(\mathrm{H}^{+}\)) instead of hydroxide ions. The presence of different ions can affect the reduction potential of the reaction because the reaction involves the transfer of electrons and a change in the oxidation state of the elements involved. The different ionic environment affects the transfer of electrons and can change the equilibrium between the species involved in the reaction, thus leading to a difference in the reduction potential. This is why the potential of the anode reaction is different in an acidic medium compared to a basic medium.$$