When `Cr(s) + OH^(-)(aq) to Cr(OH)_(4)^(-) (aq) + H_(2)(g)` (basic solution) is balanced, the sum of the coefficients of all the reactants and products is:

Options

A. 14
B. 15
C. 17
D. 9

Explanation

To balance the reaction \( \text{Cr(s)} + \text{OH}^-(\text{aq}) \rightarrow \text{Cr(OH)}_4^-(\text{aq}) + \text{H}_2(g) \) in a basic solution, we will follow these steps:

### Step 1: Identify Oxidation States
- Chromium (Cr) in solid form has an oxidation state of 0.
- In \(\text{Cr(OH)}_4^-\), let the oxidation state of Cr be \(x\). The hydroxide ion (OH\(^-\)) has an oxidation state of -1 for oxygen and +1 for hydrogen. Thus, the total charge from 4 OH\(^-\) is -4.
- The overall charge of \(\text{Cr(OH)}_4^-\) is -1, so we have:
\[ x + (-4) = -1 \implies x = +3 \]
- Therefore, Cr is oxidized from 0 to +3.

### Step 2: Identify the Half Reactions
- **Oxidation half-reaction**:
\[ \text{Cr(s)} \rightarrow \text{Cr(OH)}_4^-(\text{aq}) + 3e^- \]
- **Reduction half-reaction**:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

### Step 3: Balance Oxygen and Hydrogen
- To balance the oxygen atoms in the oxidation half-reaction, we add water (H\(_2\)O):
\[ \text{Cr(s)} + 4\text{H}_2\text{O} \rightarrow \text{Cr(OH)}_4^-(\text{aq}) + 3e^- \]
- Now, we have 4 oxygen atoms on both sides.

### Step 4: Balance Hydrogen
- We need to balance hydrogen by adding hydroxide ions (OH\(^-\)):
\[ \text{Cr(s)} + 4\text{H}_2\text{O} + 4\text{OH}^- \rightarrow \text{Cr(OH)}_4^- + 4\text{H}_2\text{O} + 3e^- \]
- After balancing, we can simplify the equation.

### Step 5: Combine Half Reactions
- To equalize the number of electrons, we multiply the reduction half-reaction by 3:
\[ 3\text{H}_2O + 6e^- \rightarrow 3\text{H}_2 + 6\text{OH}^- \]
- Now, we can add the two half-reactions:
\[ 2\text{Cr(s)} + 8\text{OH}^- + 6\text{H}_2O \rightarrow 2\text{Cr(OH)}_4^- + 3\text{H}_2 + 6\text{OH}^- \]

### Step 6: Simplify and Finalize
- Cancel out the common terms:
\[ 2\text{Cr(s)} + 2\text{OH}^- + 3\text{H}_2O \rightarrow 2\text{Cr(OH)}_4^- + 3\text{H}_2 \]

### Step 7: Calculate the Sum of Coefficients
- Coefficients:
- Reactants: 2 (Cr) + 2 (OH\(^-\)) + 3 (H\(_2\)O) = 7
- Products: 2 (Cr(OH)\(_4^-\)) + 3 (H\(_2\)) = 5
- Total sum = 7 + 5 = 12

### Final Answer
The sum of the coefficients of all the reactants and products is **12**.

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