A 1M solution of H₂SO₄ is electrolyzed. Select right statement with respect to concentration of electrolyte and products at anode and cathode respectively. Given 
$\begin{array}{l}2{\mathrm{SO}}_4^{2-}\to {\mathrm{S}}_2{\mathrm{O}}_8^{2-}+2{\mathrm{e}}^{-};{\mathrm{E}}^{\mathrm{o}}=-2.01\mathrm{V}\\ {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\to 2{\mathrm{H}}_{\text{(aq) }}^{+}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+2{\mathrm{e}}^{-};{\mathrm{E}}^{\mathrm{o}}=-1.23\mathrm{V}\end{array}$

$$\begin{gathered} Cathode :\hspace{0.17em}2H^{+} +\hspace{0.17em}2e^{-} \to H_2. \\ As E^{o }of S{O}_4^{2- }and S_2{O}_8^{2-} less than H_{2}O/O_2. \\ So the reactios at anode are \\ {H}_{2}O\hspace{0.17em}\to 2H^{+}\hspace{0.17em}+\hspace{0.17em}\frac{1}{2}{O}_2 +\hspace{0.17em}2e^{-}. \end{gathered}$$

Hence , H⁺ remain constant so H₂SO₄ concentration remains constant and the products at anode and cathode are O₂ and H₂ respectively.