Question

The standard emf for cell, Cd$_{(s)}$ | Cd$^{+2}$(1M) || Cu$^{+2}$(1M) | Cu$_{(s)}$ is 0.74 V.
If concentration of Cd$^{+2}_{(aq)}$ and Cu$^{+2}_{(aq)}$ decreases by 10 times at 298 K. Calculate emf of cell.

• A. +0.074 V
• B. +0.850 V
• C. +0.680 V
• D. +0.740 V

Hint:

If the concentration of both anode and cathode ions changes by the same factor in a symmetrical cell, the cell EMF remains equal to the standard EMF.

Answer 1

The Correct Option is D

Step 1: Nernst Equation
$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.
Step 2: Concentration Change
New $[Cd^{2+}] = 0.1$ M and new $[Cu^{2+}] = 0.1$ M.
Step 3: Substitution
$E_{cell} = 0.74 - \frac{0.0592}{2} \log \frac{0.1}{0.1} = 0.74 - 0.0296 \log(1)$.
Since $\log(1) = 0$, $E_{cell} = 0.74$ V.
Step 4: Conclusion
The emf remains 0.740 V.
Final Answer:(D)