The hydrogen electrode is dipped in a solution of pH = 3 at 25 ° C...

Given,
pH =$3$

$⇒ - log [ {H}^{+} ] = 3 [ {H}^{+} ] = {10}^{- 3}$

The half cell reaction is

$2{H}_{( aq . )}^{+} + 2{e}^{-} \to {{H}_{2}}_{\left(g\right)}$

We know,

the Nernst equation is
${E}_{red} = {E}^{0}red + 0 . 059log [ {H}^{+} ] = 0 + 0 . 059log{10}^{- 3} = + 0 . 059 \times ( - 3 ) = - 0 . 177V$

$= - 17 . 7 \times {10}^{- 2}V$.