For the cell reaction- Ni - Ni2- - - Ag- AgCalculate the equilibrium constant at 25-C- How much maximum work would be obtained by the operation of this cell-

E=E^∘ 0.0592/nlog[P]/[R] at eq^m E=0 log K=nE^∘/0.0592 Ni→ Ni^2++2e^ Ag^++ e^ → Ag E_cell^∘=0.80+0.26 =1.06 V log K=2× ...

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Equilibrium Constant from Nernst Equation

For the cell ...

Question

For the cell reaction,
$N i | N i^{2 +} | | A g^{+} | A g$
Calculate the equilibrium constant at $25^{\circ} C$ . How much maximum work would be obtained by the operation of this cell?

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Solution

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M g | M g^{2 +} (a q .) ∥ A g^{+} (a q .) | A g

25^{\circ} C

E_{M g^{2 +} / M g}^{\circ} = - 2.37 v o l t

E_{A g^{+} / A g}^{\circ} = + 0.80 v o l t

M g | M g^{2 +} | | A g^{2 +} | A g

25^{\circ} C

E_{M g / M g^{2 +}}^{\circ} = 2.37 v o l t

E_{A g^{2 +} / A g}^{\circ} = 0.80 v o l t

M g (s) | M g^{2 +} (a q) | | A g^{+} (a q) | A g (s)

25^{o}

E_{M g / M g^{2 +}}^{o} = + 2.37 V

E_{A g^{+} / A g}^{o} = 0.80 V

N i | N i^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g

(E^{\circ}

N i^{2 +} / N i = - 0.25 v o l t, E^{\circ}

A g^{+} / A g = 0.80 v o l t)

N i (s) | N i^{2 +} (a q . 0.1 M) | | A g^{+} (a q . 0.1 M | A g (s)

E_{A g^{+} / A g}^{\circ} = 0.80 v o l t

E_{N i^{2 +} / N i}^{\circ} = - 0.25 v o l t

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