Question

A voltaic cell consists of an ${\mathbf{\text{Mn/Mn}}}^{\mathbf{\text{2 +}}}$half-cell and a ${\mathbf{\text{Pb/Pb}}}^{\mathbf{\text{2 +}}}$ half-cell. Calculate $\left[{\text{Pb}}^{\text{2 +}}\right]$ when $\left[{\text{Mn}}^{\text{2 +}}\right]$ is1.44M and${{\mathit{E}}^{\mathbf{\circ }}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}$is 0.44V.

Short answer

The concentration of $P{b}^{2+}$ions is $0.345\times {10}^{20 } \text{M}$.

Step-by-step solution

Standard cell potential (Ecell).

The standard cell potential or ${\text{E}}_{\text{cell}}$is defined as the cell potential when the concentration of the species in solution is 1M. Mathematically it can be calculated as the difference of the electrode half-cell.

${\text{E}}_{\text{cell}}°={\text{E}}_{\text{cathode}}°{\text{- E}}_{\text{anode}}°$

Nernst equation

The Nernst equation establishes a relation between the cell potential under non-standard condition and concentration of the species in the solution. The Nernst equation is given below.

${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}$

Identifying anode and cathode and calculating the cell potential

We will consult appendix D for the electrode potential of the half-cell.

• Electrode potential of $Mn/M{n}^{2+}={E^{\circ }}_{Mn/M{n}^{2+}}=-1.18V$.
• Electrode potential of $P{b}^{2+}/Pb={E^{\circ }}_{P{b}^{2+}/Pb}=-0.13V$.

Since, ${E^{\circ }}_{P{b}^{2+}/Pb}>{E^{\circ }}_{Mn/M{n}^{2+}}$, ${\text{Pb}}^{\text{+ 2}}\text{/Pb}$will be cathode and ${\text{Mn/Mn}}^{\text{2 +}}$will be anode.

We will write down the half-cell reaction and calculate the standard cell potential,

$$\begin{gathered} \text{Cathode} \text{reaction} {\text{Pb}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}\to \text{Pb}\left(\text{s}\right) \\ \text{Anode} \text{reaction} \text{Mn}\left(\text{s}\right)\to \text{Mn}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}} \end{gathered}$$

Overall reaction_{${\text{Pb}}^{2+}\left(\text{aq}\right)+\text{Mn}\left(\text{s}\right)\to \text{Pb}\left(\text{s}\right)+{\text{Mn}}^{2+}\left(\text{aq}\right)$.}

_{$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{reduction}-{E^{\circ }}_{oxidation} \\ {E^{\circ }}_{cell}={E^{\circ }}_{P{b}^{2+}/Pb}-{E^{\circ }}_{M{n}^{2+}/Mn} \\ {E^{\circ }}_{cell}=-0.13V+1.18V \\ {E^{\circ }}_{cell}=1.05V \end{gathered}$$}

Using the Nernst equation to find  [Pb2+]

We know that,
$$\begin{gathered} E_{cell}={E^{\circ }}_{cell}-\frac{0.0592}{n}\log Q \\ 0.44=1.05-\frac{0.0592}{2}\log \frac{\left[M{n}^{2+}\right]}{\left[P{b}^{2+}\right]} \\ 0.44-1.05=0.0296\log \frac{1.4M}{\left[P{b}^{2+}\right]} \\ -0.61=0.0296\log \frac{1.4M}{\left[P{b}^{2+}\right]} \\ 20.608=\log \frac{1.4M}{\left[P{b}^{2+}\right]} \\ {10}^{20.603}=4.05\times {10}^{20}=\frac{1.4M}{\left[P{b}^{2+}\right]} \\ \left[P{b}^{2+}\right]=\frac{1.4M}{4.05\times {10}^{20}} \\ \left[P{b}^{2+}\right]=0.345\times {10}^{-20} \end{gathered}$$

Hence _{$\left[P{b}^{2+}\right]=0.345\times {10}^{-20}$}