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At what pH , given half cell MnO 4 - ( 0 . 1 M ) ∣ Mn 2 + ( 0 . 001 M ) will have electrode potential of 1 . 282 V ? (Nearest Integer) Given E MnO 4 - / Mn 2 + o = 1 . 54 V , 2 . 303 RT F = 0 . 059 V

Correct answer

0

Step-by-step solution

MnO 4 - + 8 H + + 5 e - ⇌ Mn 2 + + 4 H 2 O E = E ° - 0 . 059 5 log Mn 2 + MnO 4 - H + 8 1 . 282 = 1 . 54 - 0 . 059 5 log 10 - 3 10 - 1 × H + 8 0 . 258 × 5 0 . 059 = log 10 - 2 H + 8 ⇒ 21 . 86 = - 2 + 8 pH pH = 2 . 98 ≃ 3

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