Designing a galvanic cell from a single-displacement redox reaction
Suppose the galvanic cell sketched below is powered by the following reaction:
Mg(s) + ZnCl2(aq) ? MgCl2(aq) + Zn(s)
Write a balanced equation for the half-reaction that happens at the cathode of this cell.
Zn2+(aq) + 2e- ? Zn(s)
The Correct Answer and Explanation is:
Correct Answer:
Half-reaction at the cathode:
$$
\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)
$$
Detailed Explanation (300+ words):
A galvanic cell (also called a voltaic cell) is a type of electrochemical cell that uses a spontaneous redox reaction to generate electrical energy. The cell is composed of two half-cells: one for oxidation and one for reduction. The two electrodes are connected by a wire (for electron flow) and a salt bridge (for ion movement), allowing a redox reaction to occur in separate locations while producing an electric current.
In the given reaction:
$$
\text{Mg}(s) + \text{ZnCl}_2(aq) \rightarrow \text{MgCl}_2(aq) + \text{Zn}(s)
$$
This is a single-displacement redox reaction where magnesium metal (Mg) displaces zinc ions (Zn²⁺) from solution, forming magnesium ions (Mg²⁺) and solid zinc metal (Zn).
To understand what’s happening at the cathode, we must break the full reaction into half-reactions, separating the oxidation and the reduction processes.
Step 1: Identify Oxidation and Reduction
- Oxidation (loss of electrons) happens when magnesium goes from solid to ion: $$
\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-
$$ - Reduction (gain of electrons) happens when zinc ions gain electrons to form solid zinc: $$
\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)
$$
Step 2: Determine the Cathode Reaction
In a galvanic cell:
- Oxidation occurs at the anode
- Reduction occurs at the cathode
Since zinc ions are being reduced to solid zinc, this reaction takes place at the cathode.
Thus, the half-reaction at the cathode is:
$$
\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)
$$
This matches the correct answer.
Conclusion:
In this galvanic cell, magnesium is the anode (oxidized), and zinc is the cathode (reduced). The cathode half-reaction is the reduction of Zn²⁺ ions to Zn metal:
$$
\boxed{\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)}
$$
This is the correct and balanced half-reaction at the cathode.