$\frac{{4.5}}{9} = $ eq of ${H_2}$ $2{H^ + } + 2{e^ - } \to {H_2}$ eq. of ${H_2}$ =
Number of moles $× n$ factor $0.5 = \,{n_{{H_2}}} \times 2$
${V_{{H_2}}} = \frac{{0.5}}{2} \times 22.4$; ${V_{{H_2}}} = 5.6\;L$
$[Fe(CN)_6]^{4-} \rightarrow [Fe(CN)_6]^{3-} + e^{-1}\, ;$ $ E^o = -0.35\, V$
$Fe^{2+} \rightarrow Fe^{3+} + e^{-1}\ ;$ $E^o = -0.77\, V$
$Zn\,(s)\,\, + \,\,C{u^{2 + }}\,(aq)\, \rightleftharpoons \,Z{n^{2 + \,}}\,(aq)\, + Cu\,(s)$
$(R = 8 \,JK^{-1}\,mol^{-1},\, F = 96000\,C\,mol^{-1})$