Question

Answer the following questions
Calculate the standard free energy change for the following reaction at $$ 25^{\circ}\,C $$.
$$ Au(s) + Ca^{2+} (1 \, M) \rightarrow Au^{3+}(1\,M) + Ca(s) $$
$$ E^{o}_{Au^{3+} / Au} = + 1.50 \,V , E^{o}_{Ca^{2+} / Ca} = - 2.87\,V $$
Predict whether the reaction will be spontaneous or not at $$ 25^{\circ}\,C $$ . Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent ?

Answer 1

$$ E^{o}_{cell} = E^{o}_{Ca^{2+} / Ca} - E^{o}_{Au^{3+} / Au} $$
$$ = (-2.87 \,V) - (1.50\,V) = -4.37\,V $$
$$ \Delta_{r}G^{o}_{cell} = -6 \times 96500 \times (-4.37\,V) $$
$$ = + 2530.230\,kJ/mol $$
Since $$ \Delta_{r}G^{o} $$ is positive , therefore , reaction is non-spontaneous.
$$ Au^{3+} / Au $$ half cell will be an oxidizing agent while $$ Ca^{2+} / Ca $$ half cell will be a reducing agent.