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As E°cell = 0.0591/n log Kc. 0.46 = 0.0591/2 log Kc. log Kc = 2×0.46/0.0591 = 15.57 or Kc = Antilog 15.57 = 3.7×10^15 ≈ 4×10^15.

1. Kc = 4×10^15
2. Kc = 3×10^15
3. Kc = 2×10^15
4. Kc = 1×10^15

Correct answer: Kc = 4×10^15

Solution

The calculation is correct: substituting E°cell = 0.46 and n = 2 into the formula gives log Kc = 15.57, and taking the antilog results in Kc ≈ 4×10^15.

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