Question

For the cell
\(\text{Ni}_{(s)}|\text{Ni}_{(1M)}^{++}||\text{Cu}_{(1M)}^{++}|\text{Cu}_{(s)}\), if the concentration of an ion in the anode compartment is doubled, then the emf of cell will be

• A. increased by \(2\) V.
• B. decreased by \(2\) V.
• C. increased by \(0.00891\) V.
• D. decreased by \(0.00891\) V.

Hint:

Write the Nernst equation for the Ni-Cu cell; doubling the Ni2+ concentration raises the reaction quotient and lowers the emf.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The cell reaction is \(\text{Ni}+\text{Cu}^{2+}\rightarrow\text{Ni}^{2+}+\text{Cu}\), with \(n=2\). The Nernst equation is \(E=E^{0}-\dfrac{0.0591}{2}\log\dfrac{[\text{Ni}^{2+}]}{[\text{Cu}^{2+}]}\).

Step 2: Initial emf
With both concentrations \(1\text{ M}\), \(\log 1=0\), so \(E=E^{0}\).

Step 3: After doubling the Ni2+ concentration
\(E=E^{0}-\dfrac{0.0591}{2}\log 2=E^{0}-0.02955\times0.301\)

Step 4: Change in emf
\(E-E^{0}=-0.0089\text{ V}\). The emf decreases by about \(0.00891\text{ V}\), option (D).

Final Answer:
The emf decreases by about 0.00891 V. \[ \boxed{\text{(D) }\text{decreased by }0.00891\text{ V}} \]