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 redox 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: Balance the Atoms
First, we need to ensure that the number of atoms of each element is balanced on both sides of the equation.

- **Chromium (Cr)**: There is 1 Cr on both sides.
- **Oxygen (O)**: On the left, we have 4 from \(\text{Cr(OH)}_4^{-}\) and 1 from \(\text{OH}^{-}\), totaling 5 O. On the right, we have 4 from \(\text{CrO}_4^{2-}\) and 1 from \(\text{H}_2\text{O}\), totaling 5 O. So, O is balanced.
- **Hydrogen (H)**: On the left, we have 4 from \(\text{Cr(OH)}_4^{-}\) and 1 from \(\text{OH}^{-}\), totaling 5 H. On the right, we have 2 from \(\text{H}_2\text{O}\). To balance H, we need to add 3 \(\text{H}^+\) ions to the right side.

Now the equation looks like this:

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

### Step 2: Balance the Charges
Next, we need to balance the charges on both sides of the equation.

- **Left Side Charge**:
- \(\text{Cr(OH)}_4^{-}\) has a charge of -1.
- \(\text{OH}^{-}\) has a charge of -1.
- Total charge on the left = \(-1 - 1 = -2\).

- **Right Side Charge**:
- \(\text{CrO}_4^{2-}\) has a charge of -2.
- \(\text{H}_2\text{O}\) is neutral (0).
- \(3 \text{H}^{+}\) has a charge of +3.
- The charge from the electrons is \(-n\).
- Total charge on the right = \(-2 + 3 - n = 1 - n\).

### Step 3: Set Up the Charge Balance Equation
Now we set the charges equal to each other:

\[ -2 = 1 - n \]

### Step 4: Solve for \( n \)
Rearranging the equation gives:

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

### Conclusion
The value of \( n \) in the given redox reaction is \( 3 \).

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