The standard e.m.f of a cell, involving one electron change is found to be $$0.591$$ V at $$25^\circ$$C. The equilibrium constant of the reaction is ($$F = 96{,}500$$ C mol$$^{-1}$$; $$R = 8.314$$ J K$$^{-1}$$ mol$$^{-1}$$)

Given Data:

• Standard E.M.F ($$E^\circ$$) = $$0.591 \text{ V}$$
• Number of electrons involved ($$n$$) = $$1$$
• Temperature ($$T$$) = $$25^\circ\text{C} = 25 + 273 = 298 \text{ K}$$
• Faraday's constant ($$F$$) = $$96,500 \text{ C mol}^{-1}$$
• Gas constant ($$R$$) = $$8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$

The equilibrium constant given by the Nernst equation:

$$E^\circ = \frac{2.303 RT}{nF} \log_{10} K$$

At a standard temperature of $$25^\circ\text{C}$$ ($$298 \text{ K}$$), the factor $$\frac{2.303 RT}{F}$$ simplifies to approximately $$0.0591 \text{ V}$$:

$$\frac{2.303 \times 8.314 \times 298}{96,500} \approx 0.0591 \text{ V}$$

Substituting this value into the equation:

$$E^\circ = \frac{0.0591}{n} \log_{10} K$$

Now, substitute $$E^\circ = 0.591 \text{ V}$$ and $$n = 1$$:

$$0.591 = \frac{0.0591}{1} \log_{10} K$$
$$\log_{10} K = \frac{0.591}{0.0591}$$
$$\log_{10} K = 10$$

Taking the antilog (base 10) on both sides:

$$K = 10^{10} = 1.0 \times 10^{10}$$