Answer
0.035 V
Work Step by Step
$E=E^{\circ}-\frac{0.0257\,V}{n}\ln Q$
$E^{\circ}=0$ as the electrodes are identical.
Number of electrons transferred $n=2$
$Q=\frac{[Oxidised]}{[Reduced]}=\frac{0.080\,M}{1.2\,M}$
Then, we have
$E=0-(\frac{0.0257\,V}{2}\times\ln\frac{0.080}{1.2})=0.035\,V$
