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The emf of the cell Z n | Z n 2 + ( 0.01 M ) | | F e 2 + ( 0.001 M ) | Fe at 298 K is 0.2905 volt. Then the value of equilibrium constant for the cell reaction is

Options

  1. Ae 0.32 / 0.0295
  2. B1 0 0.32 / 0.0295
  3. C1 0 0.26 / 0.0295
  4. D1 0 0.32 / 0.0591

Correct answer

B. 1 0 0.32 / 0.0295

Step-by-step solution

Z n + F e 2 + → Z n 2 + + F e ( n = 2 ) E = E o − 0.0591 n log Q 0.02905 = E o - 0.0591 2 l o g 0.01 0.001 E o = 0.2905 + 0.0295 = 0.32 volt E o = 0.0591 n log   K eq 0.32 = 0.0591 2 log   K eq = 0.02945   log   K eq K eq = 10 0.32 / 0.0295

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