Question:medium

A hypothetical galvanic cell is A$_{(s)}$ | A$^+$(1M) || B$^{+2}$(1M) | B$_{(s)}$ and emf of cell is positive. What is the possible cell reaction?

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A spontaneous cell reaction must always have the oxidation half-reaction at the anode and the reduction half-reaction at the cathode, with balanced electrons.
Updated On: May 16, 2026
  • A$_{(s)}$ + B$^{+B}_{(aq)}$ $\rightarrow$ A$^{+1}_{(aq)}$ + B$_{(s)}$
  • 2A$_{(s)}$ + B$^{+2}_{(aq)}$ $\rightarrow$ 2A$^{+1}_{(aq)}$ + B$_{(s)}$
  • A$_{(s)}$ + 2B$^{+2}_{(aq)}$ $\rightarrow$ A$^{+1}_{(aq)}$ + 2B$_{(s)}$
  • 2A$^{+1}_{(aq)}$ + B$_{(s)}$ $\rightarrow$ 2A$_{(s)}$ + B$^{+2}_{(aq)}$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The question provides the standard cell notation for a galvanic cell and states that its EMF is positive. A positive EMF indicates a spontaneous reaction. We need to derive the balanced overall cell reaction from the notation.
Step 2: Key Formula or Approach:
The cell notation is given in the format: Anode | Anode electrolyte || Cathode electrolyte | Cathode.
- At the anode (left side), oxidation occurs.
- At the cathode (right side), reduction occurs.
We need to write the half-reactions for the anode and cathode and then combine them to get a balanced overall reaction.
Step 3: Detailed Explanation:
The given cell notation is A\(_{(s)}\) | A\(^+\)(1M) || B\(^{+2}\)(1M) | B\(_{(s)}\).
Anode (Oxidation) Half-Reaction:
The left side shows A\(_{(s)}\) being oxidized to A\(^+\).
\[ \text{A}_{(s)} \rightarrow \text{A}^{+}_{(aq)} + 1e^- \] Cathode (Reduction) Half-Reaction:
The right side shows B\(^{+2}\) being reduced to B\(_{(s)}\).
\[ \text{B}^{+2}_{(aq)} + 2e^- \rightarrow \text{B}_{(s)} \] Balancing the Overall Reaction:
To get the overall reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction.
- The anode reaction involves 1 electron.
- The cathode reaction involves 2 electrons.
To balance the electrons, we must multiply the anode half-reaction by 2.
\[ 2 \times (\text{A}_{(s)} \rightarrow \text{A}^{+}_{(aq)} + 1e^-) \quad \Rightarrow \quad 2\text{A}_{(s)} \rightarrow 2\text{A}^{+}_{(aq)} + 2e^- \] Now, add the modified anode half-reaction and the cathode half-reaction:
\[ 2\text{A}_{(s)} \rightarrow 2\text{A}^{+}_{(aq)} + 2e^- \] \[ \text{B}^{+2}_{(aq)} + 2e^- \rightarrow \text{B}_{(s)} \] \hrule \[ 2\text{A}_{(s)} + \text{B}^{+2}_{(aq)} \rightarrow 2\text{A}^{+}_{(aq)} + \text{B}_{(s)} \] The 2e\(^-\) on both sides cancel out. This is the balanced, spontaneous cell reaction since the EMF is positive.
Step 4: Final Answer:
The correct cell reaction is 2A\(_{(s)}\) + B\(^{+2}_{(aq)}\) \( \rightarrow \) 2A\(^{+}_{(aq)}\) + B\(_{(s)}\). This matches option (B) (assuming typos in the question's option rendering are corrected to A\(^+\) and B\(^{2+}\)).
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