Question

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 \(E_{cell}^{\circ }\) is \(0.592\ \text{V}\).

• A. \(1.25\times 10^{20}\)
• B. \(1.75\times 10^{20}\)
• C. \(1.50\times 10^{20}\)
• D. \(1.0\times 10^{20}\)

Hint:

Use E = (0.0592 / n) log K at 25 degrees Celsius with n = 2.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
At 25 degrees Celsius, \(E^{\circ}_{cell}=\dfrac{0.0592}{n}\log K\).

Step 2: Find n
Two electrons pass from Ni to \(\text{Cu}^{2+}\), so \(n=2\).

Step 3: Calculation
\(\log K=\dfrac{n\,E^{\circ}}{0.0592}=\dfrac{2\times 0.592}{0.0592}=20\)

Step 4: Equilibrium constant
\(K=10^{20}=1.0\times 10^{20}\)

Final Answer:
The equilibrium constant is 1.0 times 10 to the power 20. \[ \boxed{\text{(D) }1.0\times 10^{20}} \]