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Calculate the emf of the cell in which the following reaction takes place : Ni(s) + 2 Ag + (0.002 M) ⟶ Ni 2 + (0.160 M) + 2 Ag(s) Given that E cell o = 1 . 0 5 V

Correct answer

0.91

Step-by-step solution

From the given cell reaction and nernst equation. E cell = E cell o - 0 . 0 5 9 1 n log Ni 2 + Ag + 2 = 1 . 0 5 V - 0 . 0 5 9 1 2 log 0 . 1 6 0 0 . 0 0 2 2 = 1 . 0 5 - 0 . 0 5 9 1 2 log 4 × 1 0 4 = 1 . 0 5 - 0 . 0 5 9 1 2 4 . 6 0 2 1 = 1 . 0 5 - 0 . 1 4 = 0 . 9 1 V E cell = 0 . 9 1 V

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