Question

For the cell reaction, $A_{(s)}+B_{(aq)}^{2+}\rightarrow A_{(aq)}^{2+}+B_{(s)}$ if equilibrium constant of reaction is $10^{4}$ at 298 K. What is standard emf of cell?

• A. 0.0592 V
• B. 0.1184 V
• C. 0.1776 V
• D. 0.2368 V

Hint:

The constant 0.0592 comes from $\frac{2.303 RT}{F}$ at $298 \text{ K}$. It is sometimes approximated as 0.059 in calculations. Always determine '$n$' directly from the balanced ionic charges!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
At equilibrium, the cell potential ($E_{cell}$) is zero. The relationship between the standard EMF ($E^{\circ}_{cell}$) and the equilibrium constant ($K_c$) is derived from the Nernst equation.

Step 2: Formula Application:
$E^{\circ}_{cell} = \frac{0.0592}{n} \log K_c$

Step 3: Explanation:
From the reaction $A + B^{2+} \to A^{2+} + B$, the number of electrons transferred ($n$) is 2. $E^{\circ}_{cell} = \frac{0.0592}{2} \log (10^4)$ $E^{\circ}_{cell} = 0.0296 \times 4 = 0.1184$ V.

Step 4: Final Answer:
The standard EMF is 0.1184 V.