Question

Electrolysis of 50% \(H_2SO_4\) solution at high current density gives a compound 'A'. Hydrolysis of 'A' gives 'B'. The number of moles of \(O_2\) produced when 5 moles of 'B' reacts with acidified \(KMnO_4\) solution is:

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Hint:

Remember the sequence: \[ H_2SO_4 \xrightarrow{\text{electrolysis}} H_2S_2O_8 \] and \[ H_2S_2O_8 \xrightarrow{\text{hydrolysis}} H_2O_2 \] Hydrogen peroxide liberates oxygen when oxidized by acidified \(KMnO_4\).

Answer 1

The Correct Option is B

Concept: When concentrated sulfuric acid is electrolysed at high current density, peroxodisulfuric acid is formed. \[ 2H_2SO_4 \xrightarrow{\text{electrolysis}} H_2S_2O_8+2H^++2e^- \] Thus, \[ A=H_2S_2O_8 \] Hydrolysis of peroxodisulfuric acid produces hydrogen peroxide. \[ H_2S_2O_8+2H_2O \rightarrow 2H_2SO_4+H_2O_2 \] Hence, \[ B=H_2O_2 \]

Step 1: Write the reaction between hydrogen peroxide and acidified \(KMnO_4\). \[ 2MnO_4^-+5H_2O_2+6H^+ \rightarrow 2Mn^{2+}+8H_2O+5O_2 \] This equation shows that \[ 5\;mol\;H_2O_2 \rightarrow 5\;mol\;O_2 \]

Step 2: Use the given quantity of hydrogen peroxide. Given: \[ 5\;mol\;H_2O_2 \] From the balanced equation, \[ 5\;mol\;H_2O_2 \rightarrow 5\;mol\;O_2 \] Therefore, \[ \boxed{5} \] moles of oxygen are produced.