Question

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?

Short answer

The overall cell reaction of a Mercuric oxide dry-cell battery is: \(\mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s)\). The standard reduction potential for the anode reaction is \(-1.252 \mathrm{~V}\). The potential of the anode reaction is different in an acidic medium compared to a basic medium because the presence of different ions affects the transfer of electrons and equilibrium between the species involved in the reaction, thus leading to a difference in reduction potential.

Step-by-step solution

(a) Finding the overall cell reaction

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 \[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s). \]

(b) Finding the standard reduction potential for the anode reaction

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 \(-1.252 \mathrm{~V}\).

(c) Why the potential of the anode reaction is different in an acidic medium

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.$$

Key concepts

Half-Cell Reactions

In electrochemistry, understanding half-cell reactions is crucial. Each half-cell reaction represents either a reduction or an oxidation process that occurs in a specific part of a galvanic cell. In the mercuric oxide dry-cell batteries example, we have two half-cell reactions. The first involves the reduction of mercuric oxide:\[\mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(aq) \]This reaction takes place at the cathode, where electrons are gained. The second half-cell reaction involves the oxidation of zinc:\[\mathrm{Zn}(s) + 2 \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-}\]This occurs at the anode, where electrons are lost. By combining these reactions, the complete pathway of electron transfer in the cell can be understood. This step is fundamental for calculating the overall cell reaction and understanding the battery's energy conversion process.
When writing half-cell reactions, it's important to note the physical states of the substances (solid, liquid, aqueous) as they provide context to the conditions under which reactions occur. Keeping track of electrons ensures that the reactions adhere to the principle of charge conservation.

Standard Reduction Potential

The standard reduction potential, denoted as \(E^\circ\), is a measure of the tendency of a chemical species to be reduced. This value is obtained under standard conditions (25°C, 1 M concentration for solutions, and 1 atm pressure for gases) and is specific to each half-reaction. In our exercise, the given standard reduction potential for the cathode reaction, where mercuric oxide gains electrons, is \(+0.098 \mathrm{~V}\).
When calculating the overall cell potential, the standard reduction potential for the anode reaction also plays a critical role. The formula:\[ E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ} \]helps us determine the driving force for the cell's electrochemical process.

• To find the anode's \(E^\circ\), rearrange the formula: \(E_{\mathrm{anode}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{cell}}^{\circ}\).
• Plugging in our values: \(E_{\mathrm{anode}}^{\circ} = +0.098 \mathrm{~V} - 1.35 \mathrm{~V} = -1.252 \mathrm{~V}.\)
• This negative value indicates that zinc's oxidation has a strong tendency to lose electrons and thus drives the electrochemical cell.

Understanding these potentials gives insight into how readily reactions occur and helps predict if a cell configuration can generate electricity.

Overall Cell Reaction

To ascertain the overall cell reaction, one must combine the reactions from both the anode and the cathode. This determines the net chemical change occurring in the galvanic cell. In our exercise, after combining the two half-cell reactions:\[\mathrm{HgO}(s) + \mathrm{Zn}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s)\]This shows that solid mercuric oxide and zinc undergo a redox reaction, resulting in liquid mercury and zinc oxide as products. The electrons canceled themselves out, verifying that the reaction follows the principle of electron conservation.
The significance of the overall reaction lies in its ability to reveal the types of transformations taking place and the nature of substances being used and produced. Such reactions outline the usefulness and application potential of batteries in everyday life, underpinning their capability in delivering a steady voltage supply.