Question

$E_{Pb/Pb^{+2}}^{0} = -2\ V$

$E_{Pb^{+4}/Pb^{+2}}^{0} = -4\ V$

$E_{Pb/Pb^{+4}}^{0} = ?$

Answer 1

3 V

Answer 2

To solve this problem, we need to use the relationship between standard electrode potential ($E^0$) and Gibbs free energy ($\Delta G^0$), which is given by the formula:

$\Delta G^0 = -nFE^0$

where $n$ is the number of electrons transferred in the reaction and $F$ is Faraday's constant. Since $F$ is a constant, we can essentially work with $nE^0$ values.

Standard electrode potentials are conventionally given as standard reduction potentials. Let's interpret the given potentials accordingly:

1. $E_{Pb/Pb^{+2}}^{0} = -2\ V$: This notation refers to the reduction potential for the half-reaction $Pb^{+2} + 2e^- \rightarrow Pb$.

   Reaction (1): $Pb^{+2} + 2e^- \rightarrow Pb$

   $E^0_1 = -2\ V$

   $n_1 = 2$

   $\Delta G^0_1 = -n_1FE^0_1 = -(2)F(-2) = 4F$

2. $E_{Pb^{+4}/Pb^{+2}}^{0} = -4\ V$: This notation refers to the reduction potential for the half-reaction $Pb^{+4} + 2e^- \rightarrow Pb^{+2}$.

   Reaction (2): $Pb^{+4} + 2e^- \rightarrow Pb^{+2}$

   $E^0_2 = -4\ V$

   $n_2 = 2$

   $\Delta G^0_2 = -n_2FE^0_2 = -(2)F(-4) = 8F$

We need to find $E_{Pb/Pb^{+4}}^{0}$. This refers to the standard oxidation potential for the reaction $Pb \rightarrow Pb^{+4} + 4e^-$. To find this, first, let's find the standard reduction potential for the corresponding reduction reaction: $Pb^{+4} + 4e^- \rightarrow Pb$. Let's call this Reaction (3).

Reaction (3): $Pb^{+4} + 4e^- \rightarrow Pb$

$n_3 = 4$

$\Delta G^0_3 = -n_3FE^0_3 = -(4)FE^0_3$

We can obtain Reaction (3) by adding Reaction (1) and Reaction (2):

$(Pb^{+2} + 2e^- \rightarrow Pb) + (Pb^{+4} + 2e^- \rightarrow Pb^{+2})$

Summing these gives: $Pb^{+4} + 4e^- \rightarrow Pb$

Since Gibbs free energy is an extensive property, the $\Delta G^0$ for the combined reaction is the sum of the $\Delta G^0$ values of the individual reactions:

$\Delta G^0_3 = \Delta G^0_1 + \Delta G^0_2$

$\Delta G^0_3 = 4F + 8F = 12F$

Now, substitute this value into the equation for $\Delta G^0_3$:

$12F = -(4)FE^0_3$

Divide both sides by $F$:

$12 = -4E^0_3$

$E^0_3 = \frac{12}{-4} = -3\ V$

This value, $E^0_3 = -3\ V$, is the standard reduction potential for $Pb^{+4}/Pb$, i.e., $E_{Pb^{+4}/Pb}^{0} = -3\ V$.

The question asks for $E_{Pb/Pb^{+4}}^{0}$, which is the standard oxidation potential for $Pb \rightarrow Pb^{+4} + 4e^-$. The standard oxidation potential is the negative of the standard reduction potential:

$E_{Pb/Pb^{+4}}^{0} = -E_{Pb^{+4}/Pb}^{0}$

$E_{Pb/Pb^{+4}}^{0} = -(-3\ V) = 3\ V$