Question

The EMF of the cell at $298\ \text{K}$ is $Mg_{(s)}|Mg^{2+}(aq,\ 0.10\ \text{M})||Ag^{+}(aq,\ 0.001\ \text{M})|Ag_{(s)}$. (Given: $E^{0}_{cell}=3.17\ \text{V}$ and $\frac{2.303RT}{F}=0.06\ \text{V}$)

• A. 3.32 V
• B. 2.96 V
• C. 3.02 V
• D. 3.17 V
• E. 3.47 V

Hint:

Don't forget to square the concentration of $Ag^{+}$ because of the $2Ag^{+}$ stoichiometry.

Answer 1

The Correct Option is C

Step 1: Concept
Nernst Equation: $E_{cell} = E^{0}_{cell} - \frac{0.06}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^{2}}$. Here $n=2$.

Step 2: Meaning
Calculate the reaction quotient $Q$ and solve for $E_{cell}$.

Step 3: Analysis
$E_{cell} = 3.17 - \frac{0.06}{2} \log \frac{0.1}{(10^{-3})^{2}} = 3.17 - 0.03 \log(10^{5}) = 3.17 - 0.03(5) = 3.17 - 0.15 = 3.02~V$.

Step 4: Conclusion
The EMF is 3.02 V. Final Answer: (C)