Question

Consider the cell described below:
$\mathit{Z}\mathit{n}\left|Z{n}^{2+}\left(1 M\right)\right|\left|C{u}^{2+}\left(1M\right)\right|\mathit{C}\mathit{u}$.

Calculate the cell potential after the reaction has operated long enough for the [Zn²⁺] to have changed by0.20 mol/L. (Assume T = 25°C.)

Short answer

The $E_{cell}$value of the electrochemical cell is 1.097 V.

Step-by-step solution

Half-cell reaction at anode and cathode

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

$$\begin{gathered} Zn\left(s\right)\rightleftharpoons Z{n}^{+2}\left(aq\right)+2e^{-} \\ C{u}^{+2}\left(aq\right)+2e^{-}\rightleftharpoons Cu\left(s\right) \end{gathered}$$

The overall reaction can be denoted as:

$Zn\left(s\right)+C{u}^{+2}\left(aq\right)\rightleftharpoons Z{n}^{+2}\left(aq\right)+Cu\left(s\right)$

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,

$$\begin{gathered} {E^o}_{cell}=-0.34 V-(-0.76 V) \\ =1.1 V \end{gathered}$$

Expression of the reaction quotient

On expressing the reaction quotient for the chemical equation,

$$\begin{gathered} Q=\frac{\left[Z{n}^{+2}\right]}{\left[C{u}^{+2}\right]} \\ Q=\frac{1.2mol/L}{1mol/L} \\ Q=1.2 \end{gathered}$$

Expression of Nernst equation can be denoted as:

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log Q \\ {E}_{cell}={E^o}_{cell}-\frac{0.0591}{2}\log Q \\ {E}_{cell}=(1.1 V)-\frac{0.0591}{2}\log (1.2) \\ {E}_{cell}=1.097 V \end{gathered}$$