What is the value of n in the following equation :
`Cr(OH)_(4)^(-)+OH^(-) rarr CrO_(4)^(2-)+H_(2)O+n e`?

Options

A. 3
B. 6
C. 5
D. 2

Answer

To find the value of \( n \) in the given reaction:

\[ \text{Cr(OH)}_4^{-} + \text{OH}^{-} \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O} + n e^{-} \]

we will follow these steps:

### Step 1: Write the half-reaction
The half-reaction is already provided. We can see that chromium is being oxidized or reduced, and we need to balance the equation.

### Step 2: Count the number of hydrogen atoms
On the left side, we have:
- From \( \text{Cr(OH)}_4^{-} \): 4 hydrogen atoms
- From \( \text{OH}^{-} \): 1 hydrogen atom

Total on the left side = \( 4 + 1 = 5 \) hydrogen atoms.

On the right side, we have:
- From \( \text{H}_2\text{O} \): 2 hydrogen atoms

### Step 3: Balance the hydrogen atoms
To balance the hydrogen atoms, we can add \( 3 \text{H}^+ \) ions to the right side:

\[ \text{Cr(OH)}_4^{-} + \text{OH}^{-} \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O} + 3 \text{H}^+ + n e^{-} \]

Now, the total number of hydrogen atoms on the right side is \( 2 + 3 = 5 \), which balances with the left side.

### Step 4: Calculate the charges on both sides
Now we need to calculate the total charge on both sides to find \( n \).

**Left Side:**
- Charge from \( \text{Cr(OH)}_4^{-} \): \( -1 \)
- Charge from \( \text{OH}^{-} \): \( -1 \)

Total charge on the left side = \( -1 - 1 = -2 \)

**Right Side:**
- Charge from \( \text{CrO}_4^{2-} \): \( -2 \)
- Charge from \( \text{H}_2\text{O} \): \( 0 \)
- Charge from \( 3 \text{H}^+ \): \( +3 \)
- Charge from \( n e^{-} \): \( -n \)

Total charge on the right side = \( -2 + 0 + 3 - n = 1 - n \)

### Step 5: Set the charges equal
For the reaction to be balanced, the total charge on the left side must equal the total charge on the right side:

\[ -2 = 1 - n \]

### Step 6: Solve for \( n \)
Rearranging the equation gives:

\[ -n = -2 - 1 \]
\[ -n = -3 \]
\[ n = 3 \]

Thus, the value of \( n \) is \( 3 \).

### Final Answer
The value of \( n \) is \( 3 \).

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