Question:
Electrolysis of an aqueous solution of \(Na_2SO_4\) between Pt electrodes liberates a gas \(X\) at anode and gas \(Y\) at cathode. \(X\) and \(Y\) respectively are
Electrolysis of an aqueous solution of \(Na_2SO_4\) between Pt electrodes liberates a gas \(X\) at anode and gas \(Y\) at cathode. \(X\) and \(Y\) respectively are
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In electrolysis of aqueous \(Na_2SO_4\) using inert Pt electrodes, water is electrolysed: oxygen is liberated at anode and hydrogen at cathode.
Updated On: May 7, 2026
- \(H_2,\ O_2\)
- \(O_2,\ H_2\)
- \(SO_2,\ H_2\)
- \(H_2,\ SO_2\)
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The Correct Option is B
Solution and Explanation
Concept:
In electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. In aqueous \(Na_2SO_4\), water is mainly electrolysed because \(Na^+\) and \(SO_4^{2-}\) ions are difficult to discharge under these conditions.
Step 1: Identify the ions present in aqueous sodium sulphate solution. \[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \] Water is also present: \[ H_2O \]
Step 2: At the cathode, reduction takes place.
At cathode, water gets reduced more easily than \(Na^+\). Therefore: \[ 2H_2O+2e^- \rightarrow H_2+2OH^- \] So gas liberated at cathode is: \[ Y=H_2 \]
Step 3: At the anode, oxidation takes place.
At anode, water is oxidised to oxygen: \[ 2H_2O \rightarrow O_2+4H^+ +4e^- \] So gas liberated at anode is: \[ X=O_2 \]
Step 4: Therefore: \[ X=O_2,\qquad Y=H_2 \] Hence, \[ \boxed{O_2,\ H_2} \]
Step 1: Identify the ions present in aqueous sodium sulphate solution. \[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \] Water is also present: \[ H_2O \]
Step 2: At the cathode, reduction takes place.
At cathode, water gets reduced more easily than \(Na^+\). Therefore: \[ 2H_2O+2e^- \rightarrow H_2+2OH^- \] So gas liberated at cathode is: \[ Y=H_2 \]
Step 3: At the anode, oxidation takes place.
At anode, water is oxidised to oxygen: \[ 2H_2O \rightarrow O_2+4H^+ +4e^- \] So gas liberated at anode is: \[ X=O_2 \]
Step 4: Therefore: \[ X=O_2,\qquad Y=H_2 \] Hence, \[ \boxed{O_2,\ H_2} \]
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