To balance the redox equation, we will use the half-reaction method in acidic medium.

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To balance the redox equation, we will use the half-reaction method in acidic medium.

The unbalanced equation is:
$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{H}_2\text{O}$

Separate the equation into two half-reactions.
Oxidation half-reaction: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$
Reduction half-reaction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$
Balance atoms other than oxygen and hydrogen.
For the oxidation half-reaction, Fe atoms are already balanced.
$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$

For the reduction half-reaction, balance Cr atoms:
$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$
Balance oxygen atoms by adding $\text{H}_2\text{O}$ .
The oxidation half-reaction has no oxygen.
For the reduction half-reaction, add $7\text{H}_2\text{O}$ to the right side to balance the 7 oxygen atoms on the left:
$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Balance hydrogen atoms by adding $\text{H}^+$ .
The oxidation half-reaction has no hydrogen.
For the reduction half-reaction, add $14\text{H}^+$ to the left side to balance the $7 \times 2 = 14$ hydrogen atoms on the right:
$14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Balance the charge by adding electrons ( $\text{e}^-$ ).
For the oxidation half-reaction:
Left side charge: +2
Right side charge: +3
Add 1 electron to the right side:
$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$

For the reduction half-reaction:
Left side charge: $14(+1) + (-2) = +12$
Right side charge: $2(+3) + 0 = +6$
Add 6 electrons to the left side:
$6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Multiply the half-reactions by appropriate integers to make the number of electrons equal.
The oxidation half-reaction has 1 electron, and the reduction half-reaction has 6 electrons. Multiply the oxidation half-reaction by 6:
$6(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-)$
$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-$
Add the two balanced half-reactions and cancel out common terms (electrons).
$(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-)$
$+$
$(6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})$

$\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$
Verify the balance of atoms and charge.
Atoms:
Cr: Left = 2, Right = 2 (Balanced)
Fe: Left = 6, Right = 6 (Balanced)
O: Left = 7, Right = 7 (Balanced)
H: Left = 14, Right = 14 (Balanced)

Charge:
Left side: $(-2) + 6(+2) + 14(+1) = -2 + 12 + 14 = +24$
Right side: $2(+3) + 6(+3) + 0 = +6 + 18 = +24$ (Balanced)

The balanced equation is:
$\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$

The coefficients of $\text{Cr}_2\text{O}_7^{2-}$ and $\text{Fe}^{2+}$ are 1 and 6, respectively.

The final answer is $\text{A}$ .

To balance the redox equation, we will use the half-reaction method in acidic medium.

Answer

Step-by-step solution

Separate the equation into two half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ .

Balance the charge by adding electrons ( $\text{e}^-$ ).

Multiply the half-reactions by appropriate integers to make the number of electrons equal.

Add the two balanced half-reactions and cancel out common terms (electrons).

$(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-)$
$+$
$(6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})$

Verify the balance of atoms and charge.

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To balance the redox equation, we will use the half-reaction method in acidic medium.

Answer

Step-by-step solution

Separate the equation into two half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding H2O\text{H}_2\text{O}H2​O.

Balance hydrogen atoms by adding H+\text{H}^+H+.

Balance the charge by adding electrons (e−\text{e}^-e−).

Multiply the half-reactions by appropriate integers to make the number of electrons equal.

Add the two balanced half-reactions and cancel out common terms (electrons).

(6Fe2+→6Fe3++6e−)(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-)(6Fe2+→6Fe3++6e−) +++ (6e−+14H++Cr2O72−→2Cr3++7H2O)(6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})(6e−+14H++Cr2​O72−​→2Cr3++7H2​O)

Verify the balance of atoms and charge.

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Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ .

Balance the charge by adding electrons ( $\text{e}^-$ ).

$(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-)$
$+$
$(6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})$