Calculate the emf of the cell in which the following reaction takes place:

Given that = 1.05 V

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Solution

Applying Nernst equation we have:

= 1.05 − 0.02955 log 4 × 10⁴

= 1.05 − 0.02955 (log 10000 + log 4)

= 1.05 − 0.02955 (4 + 0.6021)

= 0.914 V

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Ni (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 Ag (s)

E_{(cell)}^{⊙} = 1.05 V

N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g (s)

E_{c e l l}^{o} = 1.05 V

N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g I

E^{o}

= 1.05

M g_{(s)} + 2 A g^{+} (0.0001 M) \to M g^{+ 2} (0.120 M) + 2 A g (s) if E_{c e l l}^{\circ} = 3.17 v o l t

E_{c e l l}

M g (s) + 2 A g^{+} (0.0001 M) \to M g^{2 +} (0.130 M) + 2 A g (s)

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