Q124 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Chemistry
Calculate the equilibrium constant at $25\ ^\circ\text{C}$ for the cell reaction, $\text{Ni(s)} + \text{Cu}^{2+}(1\text{M}) \rightarrow \text{Ni}^{2+}(1\text{M}) + \text{Cu(s)}$ if $E^\circ_{cell}$ is $0.592\ \text{V}$.
- A$1.25 \times 10^{20}$
- B$1.75 \times 10^{20}$
- C$1.50 \times 10^{20}$
- Correct answerD$1.0 \times 10^{20}$
Solution
The given cell reaction is $\text{Ni(s)} + \text{Cu}^{2+}(1\text{M}) \rightarrow \text{Ni}^{2+}(1\text{M}) + \text{Cu(s)}$.
The number of electrons transferred in the reaction is $n = 2$.
The relationship between the standard cell potential and the equilibrium constant $K_c$ at $25\ ^\circ\text{C}$ is given by:
$E^\circ_{cell} = \dfrac{0.0592}{n} \log K_c$
Substituting the given values:
$0.592 = \dfrac{0.0592}{2} \log K_c$
$\log K_c = \dfrac{0.592 \times 2}{0.0592}$
$\log K_c = 20$
$K_c = 10^{20} = 1.0 \times 10^{20}$
Answer: $1.0 \times 10^{20}$
The number of electrons transferred in the reaction is $n = 2$.
The relationship between the standard cell potential and the equilibrium constant $K_c$ at $25\ ^\circ\text{C}$ is given by:
$E^\circ_{cell} = \dfrac{0.0592}{n} \log K_c$
Substituting the given values:
$0.592 = \dfrac{0.0592}{2} \log K_c$
$\log K_c = \dfrac{0.592 \times 2}{0.0592}$
$\log K_c = 20$
$K_c = 10^{20} = 1.0 \times 10^{20}$
Answer: $1.0 \times 10^{20}$
Hence, the correct answer is option: 'D'.