CBSE Class 12 Chemistry 2024: question 25

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Nernst equation for the cell ($n=2$):
   $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[N{i}^{2+}]}{[A{g}^{+}{]}^2}$

   1 mark

2. Substitute:
   $E_{cell}=1.05-\frac{0.059}{2}\log \frac{0.1}{(0.01{)}^2}=1.05-0.0295\log 10^3$

   1 mark

3. $E_{cell}=1.05-0.0295\times 3=1.05-0.0885=0.9615$ V

   1 mark

Answer: $E_{cell}=0.9615$ V

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
• Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant