Question

Consider the galvanic cell based on the following half-reactions:

$$\begin{gathered} Z{n}^{2+}+2e^{-}\to Zn E^o=-0.76 V \\ F{e}^{2+}+2e^{-}\to Fe E^o=-0.44 V \end{gathered}$$
a. Determine the overall cell reaction and calculate cell.

b. Calculate $\Delta G^o$and K for the cell reaction at 25°C.

c. Calculate E_cell at 25°C when [Zn²⁺] = 0.10 M and[Fe²⁺] = 1.0 x 10^-5 M.

Short answer

(a) The value of ${E^o}_{cell}$will be 0.32 V.

(b) The value of $\Delta G^o$and K for the cell reaction at 25°C are -61.75kJ and respectively.

(c)$E_{\mathrm{cell}}=0.2 M$

Step-by-step solution

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,

$$\begin{gathered} {E^o}_{cell}=-0.44 V-(-0.76 V) \\ =0.32 V \end{gathered}$$

Calculate ΔGo and K for the cell reaction

On substituting the values,

$$\begin{gathered} \Delta G^o=-nF{E^o}_{cell} \\ =-(2 mol)(96485 C/mol)(0.32 V)\left(\frac{{10}^{-3 }kJ}{1 J}\right) \\ =-61.75 kJ \end{gathered}$$

On expressing the values,

$$\begin{gathered} \log K=\frac{n{E^o}_{cell}}{0.0591} \\ \log K=\frac{(2 mol)(0.32 V)}{0.0591} \\ K=6.74\times {10}^{10} \end{gathered}$$

Expression of reaction quotient

On expressing the reaction quotient for the chemical equation,

$$\begin{gathered} Q=\frac{\left[Z{n}^{+2}\right]}{\left[F{e}^{+2}\right]} \\ Q=\frac{0.1 M}{1\times {10}^{-5}M} \\ Q={10}^4 \end{gathered}$$

$$\begin{gathered} {E^o}_{cell}=-0.44 V-(-0.76 V) \\ {E^o}_{cell}=0.32 V \end{gathered}$$

On substituting the values,

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log Q \\ {E}_{cell}=0.32-\frac{0.0591}{2}\log \left({10}^4\right) \\ {E}_{cell}=0.32-0.030(4 \log 10) \\ {E}_{cell}=0.2 M \end{gathered}$$