`Cr(OH)_(3)+H_(2)O_(2)overset("Alkali")(to)CrO_(4)^(-2)+H_(2)O` the number of `OH^(-)` required to balance the above equation

Options

A. 1
B. 3
C. 4
D. 6

Answer

To balance the chemical equation \( \text{Cr(OH)}_3 + \text{H}_2\text{O}_2 \xrightarrow{\text{Alkali}} \text{CrO}_4^{2-} + \text{H}_2\text{O} \) and determine the number of hydroxide ions (\( \text{OH}^- \)) required, we can follow these steps:

### Step 1: Identify the Oxidation and Reduction Half-Reactions
The first step is to identify the oxidation and reduction half-reactions involved in the process.

- **Oxidation Half-Reaction:**
\[ \text{Cr(OH)}_3 \rightarrow \text{CrO}_4^{2-} \]

- **Reduction Half-Reaction:**
\[ \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} \]

### Step 2: Balance the Oxidation Half-Reaction
Next, we need to balance the oxidation half-reaction for oxygen and hydrogen.

- On the left, there are 3 oxygen atoms from \( \text{Cr(OH)}_3 \) and on the right, there are 4 oxygen atoms in \( \text{CrO}_4^{2-} \). To balance the oxygen atoms, we add 1 water molecule to the left side:
\[ \text{Cr(OH)}_3 + \text{H}_2\text{O} \rightarrow \text{CrO}_4^{2-} \]

- Now, we balance the hydrogen atoms. The left side has 5 hydrogen atoms (3 from \( \text{Cr(OH)}_3 \) and 2 from \( \text{H}_2\text{O} \)), while the right side has none. To balance the hydrogen atoms, we add 5 hydroxide ions (\( \text{OH}^- \)) to the left side:
\[ \text{Cr(OH)}_3 + \text{H}_2\text{O} + 5 \text{OH}^- \rightarrow \text{CrO}_4^{2-} + 5 \text{H}_2\text{O} \]

### Step 3: Balance the Reduction Half-Reaction
Now we balance the reduction half-reaction.

- The reduction half-reaction is:
\[ \text{H}_2\text{O}_2 + 2 \text{OH}^- \rightarrow 2 \text{H}_2\text{O} \]

### Step 4: Multiply the Half-Reactions
To combine the half-reactions, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction.

- From the oxidation half-reaction, we have 3 electrons involved.
- From the reduction half-reaction, we have 2 electrons involved.

To balance the electrons, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3.

### Step 5: Combine the Half-Reactions
After multiplying, we combine the two half-reactions:

- Oxidation (multiplied by 2):
\[ 2 \text{Cr(OH)}_3 + 2 \text{H}_2\text{O} + 10 \text{OH}^- \rightarrow 2 \text{CrO}_4^{2-} + 10 \text{H}_2\text{O} \]

- Reduction (multiplied by 3):
\[ 3 \text{H}_2\text{O}_2 + 6 \text{H}_2\text{O} + 6 \text{e}^- \rightarrow 6 \text{H}_2\text{O} + 6 \text{OH}^- \]

### Step 6: Final Balancing
Now we can combine the two reactions:
\[ 2 \text{Cr(OH)}_3 + 3 \text{H}_2\text{O}_2 + 10 \text{OH}^- \rightarrow 2 \text{CrO}_4^{2-} + 10 \text{H}_2\text{O} + 6 \text{OH}^- \]

### Step 7: Calculate the Number of \( \text{OH}^- \) Ions
To find the total number of \( \text{OH}^- \) ions required:
- From the oxidation half-reaction, we have 10 \( \text{OH}^- \).
- From the reduction half-reaction, we have 6 \( \text{OH}^- \).

Thus, the total number of \( \text{OH}^- \) ions required is:
\[ 10 - 6 = 4 \]

### Final Answer
The number of \( \text{OH}^- \) ions required to balance the equation is **4**.

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