For the red ox reaction :
`Cr_2O_(7)^(2-) +I^(-) +H^(+) rarr Cr^(3+) + I_2 + H_2O` the correct coefficients of the reactants for the balanced equation are

Options

A. `{:(Cr_(2)O_(7)^(2-),I^(-),H^(+)),(1,3,14):}`
B. `{:(Cr_(2)O_(7)^(2-),I^(-),H^(+)),(1,6,14):}`
C. `{:(Cr_(2)O_(7)^(2-),I^(-),H^(+)),(2,6,14):}`
D. `{:(Cr_(2)O_(7)^(2-),I^(-),H^(+)),(1,6,7):}`

Explanation

To balance the redox reaction given by:

\[ \text{Cr}_2\text{O}_7^{2-} + \text{I}^- + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{I}_2 + \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Identify the oxidation states
- In \(\text{Cr}_2\text{O}_7^{2-}\), chromium (Cr) has an oxidation state of +6.
- In \(\text{I}^-\), iodine (I) has an oxidation state of -1.
- In \(\text{Cr}^{3+}\), chromium has an oxidation state of +3.
- In \(\text{I}_2\), iodine has an oxidation state of 0.
- In \(\text{H}^+\), hydrogen has an oxidation state of +1.
- In \(\text{H}_2\text{O}\), oxygen has an oxidation state of -2 and hydrogen is +1.

### Step 2: Write the half-reactions
1. **Reduction half-reaction** (for Cr):
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

2. **Oxidation half-reaction** (for I):
\[ 6 \text{I}^- \rightarrow 3 \text{I}_2 + 6 \text{e}^- \]

### Step 3: Combine the half-reactions
Now we can combine the half-reactions:

\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{I}^- \rightarrow 2 \text{Cr}^{3+} + 3 \text{I}_2 + 7 \text{H}_2\text{O} \]

### Step 4: Identify the coefficients of the reactants
From the balanced equation, we can see the coefficients of the reactants:
- Coefficient of \(\text{Cr}_2\text{O}_7^{2-}\) is **1**.
- Coefficient of \(\text{I}^-\) is **6**.
- Coefficient of \(\text{H}^+\) is **14**.

### Final Answer
The correct coefficients of the reactants are:
- \(\text{Cr}_2\text{O}_7^{2-}\): 1
- \(\text{I}^-\): 6
- \(\text{H}^+\): 14