Question

What is the emf of the cell at $298\,K$ in which the following reaction takes place? \[ Ni(s) + 2Ag^+(0.002M) \rightarrow Ni^{2+}(0.04M) + 2Ag(s) \] ($E^\circ_{cell} = 1.05V$)

• A. 1.16 V
• B. 0.93 V
• C. 0.73 V
• D. 0.83 V
• E. 1.32 V

Hint:

$\log 10^n = n$ simplifies Nernst calculations quickly.

Answer 1

The Correct Option is B

Concept: Nernst equation: \[ E = E^\circ - \frac{0.059}{n} \log \frac{[products]}{[reactants]} \]

Step 1: Identify $n$
\[ n = 2 \]

Step 2: Reaction quotient
\[ Q = \frac{[Ni^{2+}]}{[Ag^+]^2} = \frac{0.04}{(0.002)^2} = 10000 \]

Step 3: Apply Nernst equation
\[ E = 1.05 - \frac{0.059}{2} \log(10^4) \] \[ = 1.05 - \frac{0.059}{2} \times 4 = 1.05 - 0.118 = 0.93V \] Final Conclusion:
Option (B)