Question

Write the half-reactions and the balanced cell reaction for the following galvanic cells. $$ \begin{array}{l} \text { (a) } \mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q) \| \mathrm{Cr}^{3+}(a q)\right| \mathrm{Cr}(s) \\ \text { (b) } \mathrm{Pb}(s), \mathrm{PbSO}_{4}(s) \mid \mathrm{HSO}_{4}^{-}(a q) \| \\ \mathrm{H}^{+}(a q), \mathrm{HSO}_{4}^{2-}(a q) \mid \mathrm{PbO}_{2}(s), \mathrm{PbSO}_{4}(s) \\ \text { (c) } \mathrm{Mg}(s)\left|\mathrm{Mg}^{2+}(a q) \| \mathrm{Sn}^{2+}(a q)\right| \mathrm{Sn}(s) \end{array} $$

Short answer

(a) Oxidation: \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\), Reduction: \(\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s)\), Cell Reaction: \(3\text{Zn}(s) + 2\text{Cr}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Cr}(s)\). (b) Oxidation: \(\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-\), Reduction: \(\text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2H_2O(l)\), Cell Reaction combined directly. (c) Oxidation: \(\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-\), Reduction: \(\text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s)\), Cell Reaction: \(\text{Mg}(s) + \text{Sn}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Sn}(s) \).

Step-by-step solution

Identify and Write the Oxidation Half-Reactions

For each galvanic cell, identify the anode where oxidation occurs, and write the half-reaction. For (a), the oxidation reaction is at the Zn electrode: \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \). In (b), the anode is Pb in the presence of \( \text{PbSO}_4 \) and the reaction is: \( \text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^- \). For (c), the reaction at the magnesium electrode is: \( \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \).

Identify and Write the Reduction Half-Reactions

Next, identify the cathode where reduction occurs, and write the half-reaction. For cell (a), the reduction occurs on the Cr electrode: \( \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \). For cell (b), the reaction on the cathode involving \( \text{PbO}_2 \) is: \( \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2H_2O(l)\). In cell (c), tin gets reduced: \( \text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s) \).

Balance the Electrons and Combine the Half-Reactions

For a balanced cell reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. In (a), we multiply the Zn half-reaction by 3, and the Cr half-reaction by 2 to equalize the electrons: \(3\text{Zn}(s) + 2\text{Cr}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Cr}(s)\). For (b), the half-reactions already have 2 electrons each and can be combined directly. For (c), the reactions already balance with 2 electrons each: \(\text{Mg}(s) + \text{Sn}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Sn}(s) \).

Key concepts

Oxidation Half-Reaction

Understanding the oxidation half-reaction is crucial to grasping the fundamentals of galvanic cells. In simple terms, oxidation refers to the loss of electrons by a species. During this process, the oxidation state of the substance increases. In the context of a galvanic cell, the oxidation occurs at the anode, and this is where the metal or element donates electrons and transforms from its neutral form into a positively charged ion.

For example, looking at the galvanic cell (a) from our exercise, zinc (Zn) is oxidized to zinc ions (\text{Zn}^{2+}) as it loses two electrons: \
\(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \).

This half-reaction can be visualized as zinc atoms on the surface of the metal releasing their outer electrons into the electrolyte solution, which then circulate to the reduction site or cathode.

Reduction Half-Reaction

The reduction half-reaction is the opposite process of oxidation. It involves the gain of electrons and occurs at the cathode in a galvanic cell. During reduction, a positive ion accepts electrons and becomes a neutral atom or molecule, thereby reducing its oxidation state. In cell (a), chromium ions (\text{Cr}^{3+}) are the recipient of the electrons lost by zinc:
\(\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s)\).

In a galvanic cell setup, the flow of electrons that are released by the oxidation at the anode is directed towards the cathode, where they are then accepted by the species undergoing reduction, thus completing the circuit and allowing the continued flow of current.

Electron Balance

In any chemical reaction, the law of conservation of charge must be upheld, which means the number of electrons lost and gained must be balanced. That's where the electron balance comes into play. When writing the half-reactions in a galvanic cell, it is essential to ensure that the electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction.

For the electron balances to be correct, we might have to multiply the half-reactions by appropriate coefficients. In cell (a), for example, the zinc half-reaction releases two electrons, while the chromium half-reaction requires three for full reduction. To balance the electron transfer, we need to multiply the zinc half-reaction by three and the chromium half-reaction by two, achieving a common multiple of six electrons for both half-reactions:
\(3(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-)\); \(2(\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s))\).

Cell Reaction Balancing

Cell reaction balancing involves combining the balanced oxidation and reduction half-reactions to formulate the overall cell reaction. Doing this correctly requires that both half-reactions be properly balanced in terms of both atoms and charges before you combine them. This ensures that the resulting equation represents the actual chemical changes occurring in the galvanic cell.

In our example for cell (a), after we have balanced the electrons, we combine the zinc and chromium half-reactions to get the full reaction:
\(3\text{Zn}(s) + 2\text{Cr}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Cr}(s)\).

This balanced reaction clearly indicates that three atoms of zinc are oxidized while two chromium ions are reduced, maintaining the conservation of charge and mass.