$$ E^{o}_{cell} = E^{o}_{Ca^{2+} / Ca} - E^{o}_{Au^{3+} / Au} $$
$$ = (-2.87 \,V) - (1.50\,V) = -4.37\,V $$
$$ \Delta_{r}G^{o}_{cell} = -6 \times 96500 \times (-4.37\,V) $$
$$ = + 2530.230\,kJ/mol $$
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.