Choose the set coefficients that correctly balances the following equation :
`xCr_(2)O_(7)^(2-)+yH^(+)+ze^(-)rarraCr^(+3)++bH_(2)O`

Options

A. `{:(x,y,z,a,b),(2,14,6,2,4):}`
B. `{:(x,y,z,a,b),(1,14,6,2,7):}`
C. `{:(x,y,z,a,b),(2,7,6,2,7):}`
D. `{:(x,y,z,a,b),(2,7,6,1,7):}`

Answer

To balance the redox reaction given in the question, we will follow these steps:

### Step 1: Identify the half-reaction
The reaction involves the dichromate ion \( \text{Cr}_2\text{O}_7^{2-} \) being reduced to \( \text{Cr}^{3+} \).

### Step 2: Write the unbalanced half-reaction
The unbalanced half-reaction can be written as:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \]

### Step 3: Balance the chromium atoms
There are 2 chromium atoms in the dichromate ion, so we need to have 2 chromium ions on the product side:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2 \text{Cr}^{3+} \]

### Step 4: Balance the oxygen atoms
The dichromate ion has 7 oxygen atoms. To balance the oxygen, we can add 7 water molecules to the product side:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 5: Balance the hydrogen atoms
Now, we have 14 hydrogen atoms from the 7 water molecules. To balance the hydrogen, we add 14 hydrogen ions (\( \text{H}^+ \)) to the reactant side:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 6: Balance the charges
Now we need to balance the charges. The left side has a total charge of:
- From \( \text{Cr}_2\text{O}_7^{2-} \): -2
- From \( 14 \text{H}^+ \): +14
- Total charge on the left: \( -2 + 14 = +12 \)

On the right side, the charge is:
- From \( 2 \text{Cr}^{3+} \): \( 2 \times +3 = +6 \)
- Total charge on the right: +6

To balance the charges, we need to add 6 electrons to the left side:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 7: Write the final balanced equation
Now we can summarize the coefficients:
- \( x = 1 \) (for \( \text{Cr}_2\text{O}_7^{2-} \))
- \( y = 14 \) (for \( \text{H}^+ \))
- \( z = 6 \) (for \( e^- \))
- \( a = 2 \) (for \( \text{Cr}^{3+} \))
- \( b = 7 \) (for \( \text{H}_2\text{O} \))

Thus, the balanced equation is:
\[ 1 \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Final Answer
The coefficients are:
- \( x = 1 \)
- \( y = 14 \)
- \( z = 6 \)
- \( a = 2 \)
- \( b = 7 \)