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 Question:
We are given a generic redox cell reaction and its equilibrium constant ($K_{\text{eq}}$). We must calculate the standard electromotive force (EMF or $E^\circ_{\text{cell}}$) of the cell at standard temperature (298 K).

Step 2: Detailed Explanation:
The fundamental Nernst relationship that connects the standard cell potential to the equilibrium constant is:
$E^\circ_{\text{cell}} = \frac{0.0592}{n} \log(K_{\text{eq}})$ (at $298 \text{ K}$)
1. Determine the number of transferred electrons ($n$):
Look at the balanced redox reaction: $A_{(s)} + B_{(aq)}^{2+} \rightarrow A_{(aq)}^{2+} + B_{(s)}$.
The solid metal $A$ is oxidized to $A^{2+}$, meaning it physically loses 2 electrons.
The ion $B^{2+}$ is reduced to solid metal $B$, meaning it gains those exact 2 electrons.
Therefore, the number of moles of electrons transferred in the balanced equation is $n = 2$.
2. Calculate the potential:
We are given the equilibrium constant $K_{\text{eq}} = 10^4$.
Substitute these values into the formula:
$E^\circ_{\text{cell}} = \frac{0.0592}{2} \times \log(10^4)$
Since $\log(10^4) = 4$:
$E^\circ_{\text{cell}} = \frac{0.0592}{2} \times 4$
$E^\circ_{\text{cell}} = 0.0592 \times \left(\frac{4}{2}\right)$
$E^\circ_{\text{cell}} = 0.0592 \times 2$
$E^\circ_{\text{cell}} = 0.1184 \text{ V}$

Step 3: Final Answer:
The standard emf of cell is 0.1184 V, matching option (b).