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What is the potential of an electrode which originally contained 0.1 M NO 3 - and 0.4 MH + and which has been treated by 60% of the cadmium necessary to reduce all the NO 3 - to NO(g) at 1 atm. Given, NO 3 - + 4 H + + 3 e - → NO + 2 H 2 O , E ∘ = 0.95 V and log2 = 0.3010

Options

  1. A0.52 V
  2. B0.44 V
  3. C0.86 V
  4. D0.78 V

Correct answer

C. 0.86 V

Step-by-step solution

After addition of Cd it oxidises into Cd 2+ NO 3 - aq + 4 H + aq + 3 e - → NO g + 2 H 2 O l 0.1 - x 0.4 - 4 x X = 0.06 NO 3 - remaining = 0.1 - 0.06 ≈ 0.04 M H + remaining = 0.4 - 4 × 0.06 = 0.4 - 0.24 = 0.16 M E NO 3 - / NO = E NO 3 - / NO ∘ - 0.591 3 log 1 NO 3 - H + 4 = 0.95 - 0.0591 3 log 1 0.04 0.16 4 = 0.95 - 0.0591 3 log 1 4 × 1 0 - 2 × 1 0 - 8 × 1 6 4 = 0.95 - 0.0591 3 log 1 4 × 1 0 - 2 × 1 0 - 8 × 2 + 1 6 = 0.95 - 0.0591 3 log 1 0 1 0 2 2 × 2 1 6 =

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