From the given information, we can calculate the number of moles of electrons:
\(\begin{aligned}{{}{}}{I = 2.345{\rm{A}} = 2.345\frac{{\rm{C}}}{{\rm{s}}}}\\{{\rm{t}} = 45{\rm{min}} = 45{\rm{min}} \cdot \frac{{60{\rm{s}}}}{{1{\rm{min}}}} = 2700{\rm{s}}}\end{aligned}\)
\(\begin{aligned}{{}{}}{{\rm{Q}} = I \cdot {\rm{t}} = {\rm{n}} \cdot F}\\{I \cdot {\rm{t}} = n \cdot F}\\{n = \frac{{I \cdot {\rm{t}}}}{F}}\end{aligned}\)
\({\rm{n}} = \frac{{2.345\frac{{\rm{C}}}{{\rm{s}}} \cdot 2700{\rm{s}}}}{{96485\frac{{\rm{C}}}{{{\rm{mol}} \cdot {{\rm{e}}^ - }}}}} = 0.0656{\rm{mol}}{{\rm{e}}^ - }\)
If we write the cathode reaction of water electrolysis we can see how many mole \({e^ - }\)are needed to produce \(1\)mole of \({{\rm{H}}_2}\).
Cathode:
\(2{{\rm{H}}^ + } + 2{{\rm{e}}^ - } \to {{\rm{H}}_2}\)
Now using stoichiometry, we can calculate how much \({{\rm{H}}_2},{\rm{ }}0.0656\)mole \({{\rm{e}}^ - }\)can produce:
\(2{\rm{mol}}{e^ - }:1{\rm{molH}}2 = 0.0656{\rm{mol}}{e^ - }:xmol{{\rm{H}}_2}\)
\(\begin{aligned}{{}{}}{{\rm{xmol}}{{\rm{H}}_2} = \frac{{1{\rm{mol}}{{\rm{H}}_2} \cdot 0.0656{\rm{mol}}{{\rm{e}}^ - }}}{{2{\rm{mol}}{{\rm{e}}^ - }}}}\\{xmol{H_2} = 0.0328{\rm{mol}}}\end{aligned}\)
Using the ideal gas equation, we can calculate the volume of \({{\rm{H}}_2}\).
\(\begin{aligned}{{}{}}{P \cdot {\rm{V}} = {\rm{n}} \cdot {\rm{R}} \cdot {\rm{T}}}\\{{\rm{V}} = \frac{{{\rm{n}} \cdot R \cdot {\rm{T}}}}{{\rm{P}}}}\end{aligned}\)
\(\begin{aligned}{{}{}}{{\rm{R}} = 8.314\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}}\\{{\rm{J}} = {{\rm{m}}^3} \cdot {\rm{Pa}} = {\rm{d}}{{\rm{m}}^3} \cdot {\rm{Pa}} \cdot {{10}^3} = 1 \cdot {\rm{Pa}} \cdot {{10}^3}}\end{aligned}\)
\({\rm{V}} = \frac{{0.0328{\rm{mol}} \cdot 8.314 \cdot {{10}^3}\frac{{1 \cdot {\rm{Pa}}}}{{{\rm{K}} \cdot {\rm{mol}}}} \cdot 298.15{\rm{K}}}}{{101325{\rm{Pa}}}}\)
\(V = 0.81\)
