Electrolysis of 50% `H_(2)SO_(4)` gives

Options

A. `H_(2)O`
B. `D_(2)O`
C. `H_(2)O_(2)`
D. `H_(2)`

Answer

To solve the question regarding the electrolysis of 50% \( H_2SO_4 \), we will go through the steps involved in the process:

### Step-by-Step Solution:

1. **Understanding Electrolysis**:
Electrolysis is a chemical process that uses electricity to drive a non-spontaneous reaction. In the case of \( H_2SO_4 \), it involves the decomposition of sulfuric acid into its constituent ions.

2. **Identifying the Electrolyte**:
We have a 50% solution of \( H_2SO_4 \). This means that the solution contains a significant concentration of sulfuric acid, which dissociates into \( H^+ \) and \( SO_4^{2-} \) ions in solution.

3. **Reactions at the Anode**:
At the anode (positive electrode), oxidation occurs. The sulfate ions (\( SO_4^{2-} \)) can react with water to produce \( S_2O_8^{2-} \) (peroxydisulfate) along with \( H^+ \) ions and electrons:
\[ SO_4^{2-} + H_2O \rightarrow S_2O_8^{2-} + 2H^+ + 2e^- \]

4. **Reactions at the Cathode**:
At the cathode (negative electrode), reduction occurs. The \( H^+ \) ions gain electrons to form hydrogen gas:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

5. **Formation of Hydrogen Peroxide**:
The \( S_2O_8^{2-} \) ions can further react with water to produce hydrogen peroxide (\( H_2O_2 \)):
\[ S_2O_8^{2-} + H_2O \rightarrow 2HSO_4^{-} + H_2O_2 \]

6. **Overall Reaction**:
The overall process of electrolysis of \( H_2SO_4 \) leads to the production of hydrogen gas and hydrogen peroxide:
\[ H_2SO_4 \rightarrow H_2 + H_2O_2 \]

### Conclusion:
From the electrolysis of 50% \( H_2SO_4 \), we obtain hydrogen gas (\( H_2 \)) and hydrogen peroxide (\( H_2O_2 \)).

### Final Answer:
The products of the electrolysis of 50% \( H_2SO_4 \) are \( H_2 \) and \( H_2O_2 \).

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