Question

Calculate$\Delta G^o$ and K at 25°C for the galvanic cell reactions in Exercise 21.

Short answer

Thevalue of at 25^oC for reaction a, b, c, d, are -52.1kJ, -86.8kJ, -212.3kJ, -662kJ respectively.

Thevalue of at 25^oC for reaction a, b, c, and d are $1.37\times {10}^9$, $1.7\times {10}^{15}$, $1.7\times {10}^{37}$,${10}^{116}$ respectively.

Step-by-step solution

Galvanic reaction containing Br and Cl

The oxidation and reduction half-reaction at anode and cathode can be denoted as:

$$\begin{gathered} 2B{r}^{-}\left(aq\right)\to B{r}_2\left(aq\right)+2e^{-} \\ C{l}_2\left(g\right)+2e^{-}\to 2C{l}^{-}\left(aq\right) \end{gathered}$$

The overall reaction can be written as:

${\text{2Br}}^{\text{-}}{\text{(aq) + Cl}}_{\text{2}}\text{(aq)}\to {\text{Br}}_{\text{2}}{\text{(aq) + 2Cl}}^{\text{-}} {\text{E}}^{\text{o}}\text{= 0.27} \text{J/C}$

To calculate the value,

$$\begin{gathered} \Delta G^o=-nF{E}^o \\ \Delta G^o=-(2 mol)(96.485 C/mol)(0.27 J/C) \\ \Delta G^o=-52.1 kJ \\ \log K=\frac{n{E}^o}{0.0591} \\ K=1.37\times {10}^9 \end{gathered}$$

Galvanic reaction containing Mn and IO

The oxidation and reduction half reaction at anode and cathode can be denoted as:

$$\begin{gathered} 5I{O}^{-}\left(aq\right)+10H^{+}\left(aq\right)+10e^{-}\to 5I{O_3}^{-}\left(aq\right)+5H_{2}O\left(l\right) E^o=1.6 V \\ 2M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\to 2Mn{O_4}^{-}\left(aq\right)+16H^{+}\left(aq\right)+10e^{-} E^o=-1.51 V \end{gathered}$$

The overall reaction can be denoted as:

$2M{n}^{2+}\left(aq\right)+3H_{2}O\left(l\right)+5I{O_4}^{-}\left(aq\right)+6H^{+}\left(aq\right)\to 2MN{O_4}^{-}\left(aq\right)+5I{O_3}^{-}\left(aq\right) E^o=0.09V$

To calculate,

role="math" localid="1663745489673" $$\begin{gathered} \Delta G^o=-(10 mol)(96.485 C/mol\times 0.09 J/C) \\ =-86.8 kJ \\ \log K=\frac{n{E}^o}{0.0591} \\ K=1.7\times {10}^{15} \end{gathered}$$

Galvanic reaction with oxidation and reduction half-reaction

The half-reaction of oxidation and reduction at anode and cathode can be,

$$\begin{gathered} H_2{O}_2\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}\to 2H_{2}O\left(l\right) E^o=1.78 V \\ {H}_2{O}_2\left(aq\right)\to O_2\left(g\right)+2H{}^{+}(aq)+2e^{-} E^o=-0.68 V \end{gathered}$$

The overall reaction can be,

$2H_2{O}_2\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right) E^o=1.1 V$

Then to calculate,

$$\begin{gathered} \Delta G^o=-(2 mol)(96.485 C/mol\times 1.144 J/C) \\ \Delta G^o=-212.3 kJ \\ \log K=\frac{n{E}^o}{0.0591} \\ K=1.7\times {10}^{37} \end{gathered}$$

Galvanic reaction containing Mn and Fe

The half-reaction of oxidation and reduction at anode and cathode can be:

$$\begin{gathered} 3Mn\left(s\right)\to 3M{n}^{2+}\left(aq\right)+6e^{-} E^o=1.18 V \\ 2F{e}^{3+}\left(aq\right)+6e^{-}\to 2Fe\left(s\right) E^0=-0.036 V \end{gathered}$$

The overall reaction can be denoted as:

$2F{e}^{3+}\left(aq\right)+3Mn\left(s\right)\to 3M{n}^{2+}\left(aq\right)+2Fe\left(s\right)E^0=1.144V$

Then to calculate,

$$\begin{gathered} \Delta G^o=-\left(26mol\right)(96.485 C/mol\times 1.1J/C) \\ \Delta G^o=-662 kJ \\ \log K=\frac{n{E}^o}{0.0591} \\ K={10}^{116} \end{gathered}$$