The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`

Options

A. `Na` and `Br_(2)`
B. `Na` and `O_(2)`
C. `H_(2),Br_(2)` and `NaOH`
D. `H_(2)` and `O_(2)`

Explanation

To determine the products formed when an aqueous solution of NaBr is electrolyzed in a cell with inert electrodes, we can follow these steps:

### Step 1: Dissociation of NaBr
When NaBr is dissolved in water, it dissociates into its ions:
\[ \text{NaBr} \rightarrow \text{Na}^+ + \text{Br}^- \]

### Step 2: Identify the Electrolysis Process
In electrolysis, we have two electrodes: the cathode (negative electrode) and the anode (positive electrode). The reactions at each electrode will determine the products formed.

### Step 3: Reaction at the Cathode
At the cathode, reduction occurs. The possible species that can be reduced are \(\text{Na}^+\) ions and water. However, the reduction of water is more favorable in an aqueous solution:
\[ \text{2H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \]
Thus, at the cathode, hydrogen gas (\(H_2\)) is produced along with hydroxide ions (\(OH^-\)), which can combine with sodium ions to form sodium hydroxide (\(NaOH\)):
\[ \text{Na}^+ + \text{OH}^- \rightarrow \text{NaOH} \]

### Step 4: Reaction at the Anode
At the anode, oxidation occurs. The species that can be oxidized are \(\text{Br}^-\) ions and water. The oxidation of bromide ions is more favorable:
\[ \text{2Br}^- \rightarrow \text{Br}_2 + 2e^- \]
Thus, bromine gas (\(Br_2\)) is produced at the anode.

### Step 5: Summary of Products
From the above reactions, we can summarize the products formed during the electrolysis of aqueous NaBr:
1. Hydrogen gas (\(H_2\)) at the cathode.
2. Sodium hydroxide (\(NaOH\)) in solution.
3. Bromine gas (\(Br_2\)) at the anode.

### Final Answer
The products formed when an aqueous solution of NaBr is electrolyzed in a cell with inert electrodes are:
- Hydrogen gas (\(H_2\))
- Sodium hydroxide (\(NaOH\))
- Bromine gas (\(Br_2\))