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For the reaction of A + 2B- ---- A2+ + 2B Kc is found to be 10^12. The E0 cell is:

1. 0.708 V

2. 0.1777 V

3. 0.088 V

4. 0.354 V

From the relation between free energy and electrode potential;

$\u2206G=-nF{E}_{cell}\phantom{\rule{0ex}{0ex}}$

we know that at the equilibrium change in free energy is zero (ΔG) . hence E

_{cell}

_{ }becomes zero.

From Nernst equation,

${E}_{cell}={E}^{0}-\frac{0.0592}{n}\mathrm{log}Q$

Where Q is the reaction quotient.

As E

_{cell}is zero we have,

${E}^{0}=\frac{0.0592}{n}\mathrm{log}{K}_{c}$

Here we have replaced the Q by K

_{c}as the reaction is in equilibrium.

${E}^{0}=\frac{0.0592}{2}\mathrm{log}\left({10}^{12}\right)=0.355\approx 0.340V\phantom{\rule{0ex}{0ex}}$

Regards

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