Question

Consider the cell described below:
$\mathit{A}\mathit{l}\left|A{l}^{3+}\left(1M\right)\right|\left|P{b}^{2+}\left(1M\right)\right|\mathit{P}\mathit{b}$
Calculate the cell potential after the reaction has operated long enough for the [Al³⁺] to have changed by0.60 mol/L. (Assume T = 25°C.)

Short answer

The$E_{cell}$ value of electrochemical cell is 1.5259 V.

Step-by-step solution

Half-cell reaction at anode and cathode

The half-cell reaction at anode and cathode can be expressed as:

$$\begin{gathered} Al\left(s\right)\rightleftharpoons A{l}^{+3}\left(aq\right)+3e^{-} \\ P{b}^{+2}\left(aq\right)+2e^{-}\rightleftharpoons Pb\left(s\right) \end{gathered}$$

Cell potential of an electrochemical cell

The overall cell potential can be calculated using,

${E^o}_{cell}={E^o}_{cathode}-{E^o}_{anode}$

On substituting the values,

$$\begin{gathered} {E^o}_{cell}=-0.13 V-(-1.66 V) \\ {E^o}_{cell}=1.53 V \end{gathered}$$

Expression of reaction quotient

On expressing the reaction quotient for chemical equation,

$$\begin{gathered} Q=\frac{{\left[A{l}^{+2}\right]}^2}{{\left[P{b}^{+2}\right]}^3} \\ Q=\frac{{(1.6mol/L)}^2}{{(1mol/L)}^3} \\ Q=2.56 \end{gathered}$$

Expression of Nernst equation can be denoted as:

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log Q \\ {E}_{cell}=(1.153)-\frac{0.0591}{6}\log (2.56) \\ {E}_{cell}=1.53-0.00402 \\ {E}_{cell}=1.5259 V \end{gathered}$$