Question

A disproportionation reaction involves a substance thatacts as both an oxidizing agent and a reducing agent,producing higher and lower oxidation states of the sameelement in the products. Which of the following disproportionation reactions are spontaneous under standardconditions? Calculate$\Delta G^o$ and K at 25°C for those reactions that are spontaneous under standard conditions.

$$\begin{gathered} \left(a\right)2C{u}^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+Cu\left(s\right) \\ \left(b\right)3F{e}^{2+}\left(aq\right)\to 2F{e}^{3+}\left(aq\right)+Fe\left(s\right) \\ \left(c\right)HCl{O}_2\left(aq\right)\to Cl{O_3}^{-}\left(aq\right)+HClO\left(aq\right)\left(unbalanced\right) \end{gathered}$$
Use the half-reactions:

$$\begin{gathered} Cl{O_3}^{-}+3H^{+}+2e^{-}\to HCl{O}_2+H_{2}O E^o=1.21 V \\ HCl{O}_2+2H^{+}+2e^{-}\to KClO+H_{2}O E^o=1.65 V \end{gathered}$$

Short answer

1. The values of $\Delta G^o and K$for the given reaction are -35kJ, $1.36\times {10}^6$respectively.

b.The indication of negative sign is non-spontaneous.

c. The values of $\Delta G^o and K$for the given reaction are -85kJ,$8.1\times {10}^{14}$ respectively

Step-by-step solution

Calculation of the values for equation 1

The combination of both equations can be given as:

$$\begin{gathered} {E^0}_{cell}={E^0}_{reduction}-{E^0}_{oxidation} \\ {E^0}_{cell}=0.52-0.16 \\ {E}^0=0.36 V \end{gathered}$$

The disproportionation is spontaneous as the cell potential’s value is positive.

$$\begin{gathered} \Delta G^o=-nF{E^o}_{cell} \\ \Delta G^o=1\times 96500\times 0.36 \\ \Delta G^o=-34740 J \\ \Delta G^o=-\frac{34740}{1000} \\ \Delta G^o\approx -35 kJ \\ \ln K=\frac{35 kJ}{8.314 J/K mol\left(\frac{{10}^{-3}kJ}{1 J}\right)298K} \\ K=1.36\times {10}^6 \end{gathered}$$

Calculation of the values for equation 2

The combination of both equations can be given as:

$$\begin{gathered} {E^0}_{cell}={E^0}_{reduction}-{E^0}_{oxidation} \\ =0.44-0.77 \\ {E}^0=-1.21 V \end{gathered}$$

The indication of negative sign is non-spontaneous.

Calculation of the values for equation 3 

The combination of both equations can be given as:

$$\begin{gathered} {E^0}_{cell}={E^0}_{reduction}-{E^0}_{oxidation} \\ =0.65-1.21 \\ {E}^0=0.44 V \end{gathered}$$

The disproportionation is spontaneous as the cell potential’s value is positive.

$$\begin{gathered} \Delta G^o=-nF{E^o}_{cell} \\ \Delta G^o=-2\times 96500\times 0.44 \\ \Delta G^o=-84920 J \\ \Delta G^o=-\frac{84920}{1000} \\ \Delta G^o\approx -85 kJ \\ \ln K=\frac{85 }{0.00831\times 298} \\ K=8.1\times {10}^{14} \end{gathered}$$