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For the following electrochemical cell at 298 K, P t s H 2 g , 1 b a r H + a q , 1 M | M 4 + a q , M 2 + a q P t s E c e l l = 0.092 V when [ M 2 + a q ] [ M 4 + a q ] = 1 0 x Given: E M 4 + | M 2 + 0 = 0.151 V ; 2.303 RT F = 0.059 V The value of x is

Options

  1. A- 2
  2. B- 1
  3. C1
  4. D2

Correct answer

D. 2

Step-by-step solution

Anode: H 2 s → 2 H + + 2 e - Cathode: M n 4 + + 2 e - → M n 2 + M n 4 + + H 2 → M n 2 + + 2 H + ¯ E = E ° - 0.059 2 log 10 ⁡ M n 2 + H + 2 M n 4 + P H 2 0.092 = 0.151 - 0.059 2 log 10 ⁡ ( 1 0 x ) 0.092 = 0.151 - 0.059 2 x x = 2

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