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: Equation
Cell reaction: $Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag$

Step 2: Nernst Equation
$E_{cell} = E_{cell}^{\circ} - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Ag^{+}]^{2}}$

Step 3: Analysis
If $[Ag^{+}]$ decreases from 1 M to 0.1 M ($10^{-1}$ M): - The term $- \frac{0.0592}{2} \log (\frac{1}{(10^{-1})^{2}})$ - $= - \frac{0.0592}{2} \log (10^{2})$ - $= - \frac{0.0592}{2} \times 2 = - 0.0592~V$

Step 4: Conclusion
The emf decreases by 0.0592 V. Final Answer: (B)