The cell potential for the given cell at 298KPt | H _ 2 (g, 1bar)||...

{H}_{2}$\left(g\right)$ + {Cu}^{2 +}\left(aq .\right)$\to 2{H}^{+}\left(aq .\right)$ + Cu\left(s\right)$</p><p>Applying nernst equation</p><p>${E}_{cell} = {E}_{cell}^{0} - \dfrac{0 . 06}{n}log\dfrac{\left[products\right]}{\left[Reactants\right]}$</p><p>${E}_{cell} = Cellpotential = 0 . 31V{E}_{cell}^{0} = 0 . 34{p}^{H} = 3, Hence, \left[{H}^{+}\right] = {10}^{- 3}$</p><p>$0 . 31 = 0 . 34 - \dfrac{0 . 06}{2}log\dfrac{{\left[{H}^{+}\right]}^{2}}{\left[{Cu}^{2 +}\right]}$</p><p>\left[{Cu}^{2 +}\right] = {10}^{- 7}M$

$x = 7$