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 basic solution and find the sum of the coefficients of all the reactants and products, we can follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced equation is:
\[ \text{Cr(s)} + \text{OH}^-(\text{aq}) \rightarrow \text{Cr(OH)}_4^-(\text{aq}) + \text{H}_2(g) \]

### Step 2: Identify the half-reactions
We can break the overall reaction into two half-reactions:
1. Oxidation half-reaction: \( \text{Cr} \rightarrow \text{Cr(OH)}_4^- \)
2. Reduction half-reaction: \( \text{OH}^- \rightarrow \text{H}_2 \)

### Step 3: Balance the oxidation half-reaction
For the oxidation half-reaction:
\[ \text{Cr} \rightarrow \text{Cr(OH)}_4^- \]
- Balance the oxygen by adding water:
\[ \text{Cr} + 4\text{H}_2\text{O} \rightarrow \text{Cr(OH)}_4^- \]
- Now, balance the hydrogen by adding \( \text{H}^+ \):
\[ \text{Cr} + 4\text{H}_2\text{O} \rightarrow \text{Cr(OH)}_4^- + 8\text{H}^+ \]
- Add \( \text{OH}^- \) to both sides to neutralize \( \text{H}^+ \):
\[ \text{Cr} + 4\text{OH}^- \rightarrow \text{Cr(OH)}_4^- + 4\text{H}_2\text{O} \]
- Balance the charge by adding electrons:
\[ \text{Cr} + 4\text{OH}^- + 3\text{e}^- \rightarrow \text{Cr(OH)}_4^- + 4\text{H}_2\text{O} \]

### Step 4: Balance the reduction half-reaction
For the reduction half-reaction:
\[ \text{OH}^- \rightarrow \text{H}_2 \]
- Balance the oxygen by adding water:
\[ 2\text{OH}^- \rightarrow \text{H}_2 + \text{H}_2\text{O} \]
- Balance the hydrogen by adding \( \text{OH}^- \):
\[ 2\text{OH}^- \rightarrow \text{H}_2 + 2\text{H}^+ \]
- Add \( \text{OH}^- \) to both sides to neutralize \( \text{H}^+ \):
\[ 2\text{OH}^- + 2\text{H}_2\text{O} \rightarrow \text{H}_2 + 3\text{OH}^- \]
- Balance the charge by adding electrons:
\[ 2\text{OH}^- + 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 3\text{OH}^- \]

### Step 5: Equalize the number of electrons
To combine the half-reactions, we need to make sure the number of electrons is equal. We can multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:
- Oxidation half-reaction (multiplied by 2):
\[ 2\text{Cr} + 8\text{OH}^- + 6\text{e}^- \rightarrow 2\text{Cr(OH)}_4^- + 8\text{H}_2\text{O} \]
- Reduction half-reaction (multiplied by 3):
\[ 6\text{OH}^- + 6\text{H}_2\text{O} + 6\text{e}^- \rightarrow 3\text{H}_2 + 9\text{OH}^- \]

### Step 6: Combine the half-reactions
Now, we can combine the two half-reactions:
\[ 2\text{Cr} + 8\text{OH}^- + 6\text{e}^- + 6\text{OH}^- + 6\text{H}_2\text{O} + 6\text{e}^- \rightarrow 2\text{Cr(OH)}_4^- + 8\text{H}_2\text{O} + 3\text{H}_2 + 9\text{OH}^- \]

### Step 7: Simplify the equation
After canceling out the electrons and simplifying, we arrive at the balanced equation:
\[ 2\text{Cr} + 8\text{OH}^- \rightarrow 2\text{Cr(OH)}_4^- + 3\text{H}_2 \]

### Step 8: Calculate the sum of coefficients
Now, we can find the sum of the coefficients:
- For reactants: \( 2 + 8 = 10 \)
- For products: \( 2 + 3 = 5 \)

Thus, the total sum of coefficients is:
\[ 10 + 5 = 15 \]

### Final Answer
The sum of the coefficients of all the reactants and products is **15**.