To balance the redox equation, we will use the half-reaction method in acidic medium.
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
To balance the redox equation, we will use the half-reaction method in acidic medium.
The unbalanced equation is:
$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{H}_2\text{O}$
Step 1: Separate the equation into two half-reactions.
Oxidation half-reaction: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$
Reduction half-reaction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$
Step 2: 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:
$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$
Step 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:
$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Step 4: 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:
$14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Step 5: 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:
$6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Step 6: 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:
$6(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-)$
$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-$
Step 7: Add the two balanced half-reactions and cancel out common terms (electrons).
$(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-)$
$+$
$(6\text{e}^- + 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O})$
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$\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$
Step 8: 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 $\boxed{\text{A}}$.
