The product obtained at anode when 50% ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ aqueous solution is electrolysed using platinum electrodes is

${{\rm{H}}_2}{{\rm{S}}_2}{{\rm{O}}_8}$

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_3}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

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Understanding +4 | -1

The product obtained at anode when 50%

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

aqueous solution is electrolysed using platinum electrodes is

View answer and explanation

Correct option: B

{\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}

Option B (

{\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}

) is correct because at high concentrations of sulfuric acid, peroxydisulfuric acid (

H_2S_2O_8

) is formed at the anode.

Practice All Chapter MCQs →

A{l^{3 + }}

{H_2}

{H^ + }

\mathrm{LiCl},{\rm{NaCl {\kern 1pt}{\kern 1pt}and {\kern 1pt}{\kern 1pt}KCl}}

\left(\mathrm{CuSO}_4\right)

\mathrm{A}{\mathrm{l}^{3 + }}