Question

What happens to the emf for the cell, $Zn|Zn_{(aq)}^{2+}||Ag_{(aq)}^{+}|Ag_{(s)}$ if concentration of $Ag^{+}$ decreases to 0.1 M?

• A. increase by 0.0592 V
• B. decrease by 0.0592 V
• C. increase by 0.0296 V
• D. decrease by 0.0296 V

Hint:

Decreasing reactant concentration always decreases the cell potential.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the change in cell potential (\( \Delta \text{E}_{\text{cell}} \)) when the concentration of a reactant ion is changed, using the Nernst equation.

Step 2: Key Formula or Approach:
Overall reaction: \( \text{Zn}_{(\text{s})} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}_{(\text{s})} \); where \( \text{n} = 2 \).
\[ \text{E}_{\text{cell}} = \text{E}^{\circ}_{\text{cell}} - \frac{0.0592}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \right) \]

Step 3: Detailed Explanation:
Initial state: \( [\text{Zn}^{2+}] = 1 \text{ M, } [\text{Ag}^+] = 1 \text{ M} \).
\( \text{E}_1 = \text{E}^{\circ}_{\text{cell}} - 0 = \text{E}^{\circ}_{\text{cell}} \).
Final state: \( [\text{Zn}^{2+}] = 1 \text{ M, } [\text{Ag}^+] = 0.1 \text{ M} \).
\[ \text{E}_2 = \text{E}^{\circ}_{\text{cell}} - \frac{0.0592}{2} \log \left( \frac{1}{(0.1)^2} \right) \]
\[ \text{E}_2 = \text{E}^{\circ}_{\text{cell}} - \frac{0.0592}{2} \log(100) \]
\[ \text{E}_2 = \text{E}^{\circ}_{\text{cell}} - \frac{0.0592}{2} \times 2 = \text{E}^{\circ}_{\text{cell}} - 0.0592 \text{ V} \]
Change \( \Delta \text{E} = \text{E}_2 - \text{E}_1 = -0.0592 \text{ V} \).
This means the emf decreases by 0.0592 V.

Step 4: Final Answer:
The emf decreases by 0.0592 V.