# Will lead (Pb) react with HgSO_4?

Feb 13, 2016

Let us examine their standard reduction potentials. For mercury we have to combine two different reduction reactions.

$2 {\text{Hg"^(2+)(aq) + 2e^(-) -> "Hg}}_{2}^{2 +} \left(a q\right)$ $\text{ }$ ${E}_{\text{red"^@ = "+0.92 V}}$

$\text{Hg"_2^(2+)(aq) + 2e^(-) -> 2"Hg} \left(l\right)$ $\text{ }$ ${E}_{\text{red"^@ = "+0.85 V}}$

$\text{Pb"^(2+)(aq) + 2e^(-) -> "Pb} \left(s\right)$ $\text{ }$ ${E}_{\text{red"^@ = "-0.13 V}}$

The first two cancel out to get:

2"Hg"^(2+)(aq) + 2e^(-) -> cancel("Hg"_2^(2+)(aq)) $\text{ "" }$ ${E}_{\text{red"^@ = "+0.92 V}}$
cancel("Hg"_2^(2+)(aq)) + 2e^(-) -> 2"Hg"(l) $\text{ "" }$ ${E}_{\text{red"^@ = "+0.85 V}}$
$\text{--------------------------------------------}$
$\textcolor{g r e e n}{2 \text{Hg"^(2+)(aq) + 4e^(-) -> 2"Hg} \left(l\right)}$ $\text{ "" }$ $\textcolor{g r e e n}{{E}_{\text{red"^@ = "+1.77 V}}}$

Now, when we add it to the iron reaction, in order for the reaction to work overall, ${E}_{\text{cell}}^{\circ}$ has to be positive because of the following two equations:

$\setminus m a t h b f \left(\Delta {G}^{\circ} = - n F {E}_{\text{cell}}^{\circ}\right)$ (1)

$\setminus m a t h b f \left(\Delta G = \Delta {G}^{\circ} + R T \ln Q\right)$ (2)

where:

• $\Delta G$ is the Gibbs' free energy, and $\Delta {G}^{\circ}$ is that in standard conditions (${25}^{\circ} \text{C}$ and $\text{1 bar}$).
• $n$ is the number of $\text{mol}$s of electrons.
• $F$ is Faraday's constant, which is about $\text{96458 C/mol}$.
• ${E}_{\text{cell"^@ = E_"red"^@ + E_"oxid}}^{\circ}$ is the relationship of the standard cell potential to the standard reduction and oxidation potentials.
• $R$ is the universal gas constant in $\left(\text{C"cdot"V")/("mol"cdot"K}\right)$ ($\text{1 J = 1 C"cdot"V}$).
• $T$ is the temperature in $\text{K}$.

and it has to cancel out the electrons.

Now, since in standard conditions (${25}^{\circ} \text{C}$, $\text{1 bar}$), we assume that all concentrations are $\text{1 M}$, $Q = K$ and $\Delta {G}^{\circ} = \Delta G$.

Therefore, when ${E}_{\text{cell}}^{\circ} > 0$, $\Delta {G}^{\circ} < 0$ and $\Delta G < 0$, so the reaction would be spontaneous in standard conditions.

In this case it's not too difficult because since the reduction potential of mercury is much higher, it is easily reduced and so it makes a good oxidizing agent.

$\frac{1}{2} \left(2 \text{Hg"^(2+)(aq) + cancel(4e^(-)) -> 2"Hg} \left(l\right)\right)$ $\text{ }$ $\frac{1}{2} {E}_{\text{red"^@ = 1/2xx"+1.77 V}}$
${\text{Pb"(s) -> "Pb}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}$ $\text{ }$ ${E}_{\text{oxid"^@ = "+0.13 V}}$
$\text{--------------------------------------------}$
$\textcolor{g r e e n}{\text{Hg"^(2+)(aq) + "Pb"(s) -> "Pb"^(2+)(aq) + "Hg} \left(l\right)}$ $\text{ }$ $\textcolor{g r e e n}{{E}_{\text{cell"^@ = "+1.02 V}}}$

Therefore, add back the spectator anion to get the full reaction:

$\textcolor{b l u e}{\text{HgSO"_4(aq) + "Pb"(s) -> "PbSO"_4(aq) + "Hg} \left(l\right)}$

Now keep in mind that mercury (II) sulfate apparently decomposes in water, so this would probably have to be done in a different solvent.

I expect that the reduction potentials would change, but their relativities should remain the same (i.e. the reduction potential of mercury(II) should still remain higher than that of iron(II)).