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A 1 M solution of H2SO4 is electrolysed....

A 1 M solution of `H_2SO_4` is electrolysed. Select correct statement in respect of products obtain at anode and cathode respectively:
Given : `2SO_4^(2-) toS_2O_8^(2-)+2e^-,E^(@)=-1.23V`
`H_2O(l)to2H^+(aq)+1//2O_2(g)+2e^-,E^(@)=-1.23V`

A

concentration of `H_2SO_4` remain constant, `H_2,O_2`

B

concentration of `H_2SO_4` remain constant, `O_2,H_2`

C

concentration of `H_2SO_4` remain constant, `O_2,H_2`

D

concentration of `H_2SO_4` remain constant, `S_2O_8^(2-),H_2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the electrolysis of a 1 M solution of \( H_2SO_4 \), we need to analyze the reactions occurring at the anode and cathode during the electrolysis process. ### Step-by-Step Solution: 1. **Identify the Reactions**: - At the anode, oxidation occurs. The given reaction is: \[ H_2O(l) \rightarrow 2H^+(aq) + \frac{1}{2}O_2(g) + 2e^- \] with a standard electrode potential \( E^\circ = -1.23 \, V \). - At the cathode, reduction occurs. The relevant reaction is: \[ 2H^+(aq) + 2e^- \rightarrow H_2(g) \] with a standard electrode potential \( E^\circ = 0 \, V \) (standard hydrogen electrode). 2. **Determine the Anode Reaction**: - The oxidation of water at the anode produces oxygen gas and protons. The reaction is: \[ H_2O \rightarrow 2H^+ + \frac{1}{2}O_2 + 2e^- \] - Therefore, at the anode, the product is \( O_2(g) \). 3. **Determine the Cathode Reaction**: - The reduction of protons at the cathode produces hydrogen gas. The reaction is: \[ 2H^+ + 2e^- \rightarrow H_2(g) \] - Therefore, at the cathode, the product is \( H_2(g) \). 4. **Conclusion on Products**: - From the above reactions, we conclude that during the electrolysis of a 1 M solution of \( H_2SO_4 \): - At the anode, oxygen gas (\( O_2 \)) is produced. - At the cathode, hydrogen gas (\( H_2 \)) is produced. 5. **Consider the Concentration of \( H_2SO_4 \)**: - The concentration of \( H_2SO_4 \) remains constant during the electrolysis process as the ions are not consumed in a way that changes the overall concentration significantly. ### Final Answer: The correct statement regarding the products obtained at the anode and cathode is: - **Anode: \( O_2(g) \); Cathode: \( H_2(g) \)**.
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