Answer
Please see the work below.
Work Step by Step
As $E^{\circ}$ is less negative so $H^{+2}$ is reduced at the cathode and is written on the right, $Al$ is oxidized at the anode and is written first. The notation is given as
$Al(s)\vert Al^{+3}(aq)\vert \space \vert H^{+}(aq)\vert \space \vert H_2(gas)\vert Pt$
