For the redox reaction
`xFe^(2+)+yCr_(2)O_(7)^(2-)+zH^(+) rarr Fe^(3+)+Cr^(3+)+H_(2)O`
`x`, `y` and `z` are

Options

A. (`x=3`, `y=1`, `z=14`)
B. (`x=6`, `y=1`, `z=7`)
C. (`x=6`, `y=2`, `z=14`)
D. (`x=6`, `y=1`, `z=14`)

Explanation

To solve the redox reaction given by:

\[ x \text{Fe}^{2+} + y \text{Cr}_2\text{O}_7^{2-} + z \text{H}^+ \rightarrow \text{Fe}^{3+} + \text{Cr}^{3+} + \text{H}_2\text{O} \]

we will follow the ion-electron method to balance the reaction. Here are the steps:

### Step 1: Identify the oxidation and reduction half-reactions
- **Oxidation Half-Reaction**: Iron (Fe) is oxidized from Fe²⁺ to Fe³⁺.
- **Reduction Half-Reaction**: Dichromate ion (Cr₂O₇²⁻) is reduced to Cr³⁺.

### Step 2: Write the oxidation half-reaction
The oxidation half-reaction can be written as:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

### Step 3: Balance the oxidation half-reaction
The oxidation half-reaction is already balanced in terms of atoms and charge, as it involves the loss of one electron.

### Step 4: Write the reduction half-reaction
The reduction half-reaction can be written as:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \]

### Step 5: Balance the chromium atoms in the reduction half-reaction
Since there are 2 chromium atoms in the dichromate ion, we write:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2 \text{Cr}^{3+} \]

### Step 6: Balance the oxygen atoms
Dichromate has 7 oxygen atoms, so we need to 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 7: Balance the hydrogen atoms
Now we have 14 hydrogen atoms from the 7 water molecules, so we need to add 14 H⁺ ions 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 8: Balance the charge in the reduction half-reaction
The left side has a charge of \(2 - 14 = -12\) and the right side has a charge of \(2 \times 3 = +6\). 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 9: Combine the half-reactions
To combine the half-reactions, we need to multiply the oxidation half-reaction by 6 so that the electrons cancel out:
\[ 6 \text{Fe}^{2+} \rightarrow 6 \text{Fe}^{3+} + 6 e^- \]

Now, we can add the oxidation and reduction half-reactions:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 10: Identify the coefficients \(x\), \(y\), and \(z\)
From the balanced equation, we can see:
- \(x = 6\) (for Fe²⁺)
- \(y = 1\) (for Cr₂O₇²⁻)
- \(z = 14\) (for H⁺)

Thus, the values of \(x\), \(y\), and \(z\) are:
- \(x = 6\)
- \(y = 1\)
- \(z = 14\)

### Final Answer
The values of \(x\), \(y\), and \(z\) are \(6\), \(1\), and \(14\) respectively.

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