Balance given following half reaction for the unbalanced whole reaction :
`CrO_(4)^(2-)rarrCrO_(2)^(-)_OH^(-)` is :

Options

A. `CrO_(4)^(-2)+2H_(2)O+3e^(-)rarrCrO_(2)^(-)+4OH^(-)`
B. `2CrO_(4)^(2-)+8H_(2)OrarrCrO_(2)^(-)+4H_(2)O+8OH^(-)`
C. `CrO_(4)^(-2)+H_(2)OrarrCrO_(2)^(-)+H_(2)O+OH^(-)`
D. `3CrO_(4)^(-2)+4H_(2)O+6e^(-)rarr2CrO_(2)^(-1)+8OH^(-)`

Answer

To balance the half-reaction \( \text{CrO}_4^{2-} \rightarrow \text{CrO}_2^{-} + \text{OH}^- \) in a basic medium, we can follow these steps:

### Step 1: Write the unbalanced half-reaction
The half-reaction we need to balance is:
\[ \text{CrO}_4^{2-} \rightarrow \text{CrO}_2^{-} + \text{OH}^- \]

### Step 2: Balance the oxygen atoms
We have 4 oxygen atoms on the left side and 2 on the right side. To balance the oxygen atoms, we can add water molecules to the side that is deficient in oxygen. Since the right side is deficient, we add 2 water molecules:
\[ \text{CrO}_4^{2-} \rightarrow \text{CrO}_2^{-} + 2 \text{H}_2\text{O} \]

### Step 3: Balance the hydrogen atoms
Now, we have 4 hydrogen atoms on the right side (from 2 water molecules). To balance the hydrogen, we add hydroxide ions (\( \text{OH}^- \)) to the left side. We need to add 4 hydroxide ions:
\[ \text{CrO}_4^{2-} + 4 \text{OH}^- \rightarrow \text{CrO}_2^{-} + 2 \text{H}_2\text{O} \]

### Step 4: Balance the charges
Now, let's check the charges:
- Left side: \( 2- + 4(-1) = -6 \)
- Right side: \( -1 \)

To balance the charges, we need to add electrons to the left side. We need to add 3 electrons to the left side to make both sides equal:
\[ \text{CrO}_4^{2-} + 4 \text{OH}^- + 3 e^- \rightarrow \text{CrO}_2^{-} + 2 \text{H}_2\text{O} \]

### Step 5: Simplify the equation
We can simplify the equation by subtracting the water molecules. We have 2 water molecules on the right side and 2 on the left side (from the hydroxide ions). Thus, we can rewrite the balanced half-reaction:
\[ \text{CrO}_4^{2-} + 3 e^- + 2 \text{H}_2\text{O} \rightarrow \text{CrO}_2^{-} + 4 \text{OH}^- \]

### Final Balanced Half-Reaction
The final balanced half-reaction is:
\[ \text{CrO}_4^{2-} + 2 \text{H}_2\text{O} + 3 e^- \rightarrow \text{CrO}_2^{-} + 4 \text{OH}^- \]