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

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

[Given that $E_{(cell)}^{⊙} = 1.05 V$ ]

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Solution

$E = E^{\circ} - \frac{0.0591}{2} log \frac{[0.160]}{{[0.002]}^{2}}$

$= 1.05 - \frac{0.0591}{2} \times 4.60$

$= 1.05 - 0.136 = 0.914$

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Similar questions

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

Given that = 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)

E_{(c e l l)} if E_{(c e l l)}^{o} = 3.17 V

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