In an electrochemical reaction of lead, at standard temperature, if $$E^0_{Pb^{2+}/Pb} = m$$ Volt and $$E^0_{Pb^{4+}/Pb} = n$$ Volt, then the value of $$E^0Pb^{2+}/Pb^{4+}$$ is given by $$m - xn$$. The value of x is _______ (Nearest integer).

We are given:

$$E^0_{Pb^{2+}/Pb} = m$$ V (Pb$$^{2+}$$ + 2e$$^-$$ → Pb)

$$E^0_{Pb^{4+}/Pb} = n$$ V (Pb$$^{4+}$$ + 4e$$^-$$ → Pb)

We need to find $$E^0_{Pb^{4+}/Pb^{2+}}$$ (Pb$$^{4+}$$ + 2e$$^-$$ → Pb$$^{2+}$$).

Using Gibbs energy:

$$\Delta G_1 = -4Fn$$ (for Pb$$^{4+}$$ → Pb)

$$\Delta G_2 = -2Fm$$ (for Pb$$^{2+}$$ → Pb)

For the reaction Pb$$^{4+}$$ + 2e$$^-$$ → Pb$$^{2+}$$:

$$\Delta G_3 = \Delta G_1 - \Delta G_2 = -4Fn - (-2Fm) = -4Fn + 2Fm$$

$$E^0_{Pb^{4+}/Pb^{2+}} = -\frac{\Delta G_3}{2F} = \frac{4Fn - 2Fm}{2F} = 2n - m$$

The question asks for $$E^0_{Pb^{2+}/Pb^{4+}}$$ (oxidation potential) = $$-(2n - m) = m - 2n$$

Comparing with $$m - xn$$: $$x = 2$$