# Question #4a3ac

May 15, 2015

${E}^{\circ} = + 0.37 \text{V}$

$\Delta {G}^{\circ} = - 35.7 \text{kJ}$

${K}_{c} = 1.82 \times {10}^{6} \text{mol"^(-1)."l}$

#### Explanation:

Start with the ${E}^{\circ}$ values, listing them in increasing order:

$C {u}_{\left(a q\right)}^{2 +} + e r i g h t \le f t h a r p \infty n s C {u}_{\left(a q\right)}^{+}$ ${E}^{\circ} = + 0.153 \text{V}$ $\textcolor{red}{\left(1\right)}$

$C {u}_{\left(a q\right)}^{+} + e r i g h t \le f t h a r p \infty n s C {u}_{\left(s\right)}$ ${E}^{\circ} = + 0.52 \text{V}$ $\textcolor{red}{\left(2\right)}$

To work out ${E}_{\left(c e l l\right)}^{\circ}$ subtract the least +ve value from the most +ve.:

${E}_{\left(c e l l\right)}^{\circ} = + 0.52 - 0.153 = + 0.37 \text{V}$

This means $\textcolor{red}{\left(2\right)}$ will be driven left to right and $\textcolor{red}{\left(1\right)}$ will be driven right to left$\Rightarrow$

$2 C {u}_{\left(a q\right)}^{+} \rightarrow C {u}_{\left(a q\right)}^{2 +} + C {u}_{\left(s\right)}$

$\Delta {G}^{\circ} = - n F {E}_{\left(c e l l\right)}^{\circ}$

$= - 1 \times 9.65 \times {10}^{4} \times 0.37 = - 35705 \text{J}$

$= - 35.7 \text{kJ}$

$\Delta {G}^{\circ} = - R T \ln {K}_{c}$

$\ln {K}_{c} = \frac{- \Delta G}{R T}$

$\ln {K}_{c} = \frac{35.7 \times {10}^{3}}{8.31 \times 298} = 14.416$

From which:

${K}_{c} = 1.82 \times {10}^{6} \text{mol"^(-1)."l}$

This is an example of a disproportionation reaction where a species, in this case copper(1), is simultaneously oxidised and reduced.