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MHT CET 202625 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 25 April · Evening Shift · Chemistry · Electrochemistry

Calculate the equilibrium constant at 25\ ^ ^ C for the cell...

Calculate the equilibrium constant at 25\ ^{}^{\circ}\text{C} for the cell reaction, Ni(s)+Cu2+(1M)→Ni2+(1M)+Cu(s)\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 V0.592\ \text{V}.

Options

  1. A1.25×10201.25 \times 10^{20}
  2. B1.75×10201.75 \times 10^{20}
  3. C1.50×10201.50 \times 10^{20}
  4. D1.0×10201.0 \times 10^{20}

Correct answer

D. 1.0×10201.0 \times 10^{20}

Step-by-step solution

The given cell reaction is Ni(s)+Cu2+(1M)→Ni2+(1M)+Cu(s)\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=2n = 2.

The relationship between the standard cell potential and the equilibrium constant KcK_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=0.05922log⁡Kc0.592 = \dfrac{0.0592}{2} \log K_c

log⁡Kc=0.592×20.0592\log K_c = \dfrac{0.592 \times 2}{0.0592}

log⁡Kc=20\log K_c = 20

Kc=1020=1.0×1020K_c = 10^{20} = 1.0 \times 10^{20}

Answer: 1.0×10201.0 \times 10^{20}
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