Question

In the balanced redox reaction \(\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{SO}_3^{2-} \longrightarrow \mathrm{Cr}^{3+} + \mathrm{SO}_4^{2-}\), the ratio of the number of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) ions to \(\mathrm{SO}_4^{2-}\) ions is:

• A. \( 1:1 \)
• B. \( 1:2 \)
• C. \( 1:3 \)
• D. \( 2:3 \)

Hint:

For balancing redox reactions:
- Split into oxidation and reduction half-reactions.
- Balance atoms (O with \(\mathrm{H}_2\mathrm{O}\), H with \(\mathrm{H}^+\) in acidic medium).
- Balance charge with electrons.
- Equalize electrons and combine half-reactions.
- Check atom and charge balance in the final equation.

Answer 1

The Correct Option is C

To find the ratio of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) to \(\mathrm{SO}_4^{2-}\) in the balanced redox reaction, we need to balance the reaction in acidic medium (standard for such reactions unless specified otherwise) and compare the stoichiometric coefficients.
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Step 1: Write half-reactions
The reaction involves: - Reduction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}\) - Oxidation: \(\mathrm{SO}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}\) Reduction half-reaction:
\[ \mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+} \] - Balance chromium: Left has 2 Cr, right has 2 Cr. - Balance oxygen: Left has 7 O, so add 7 \(\mathrm{H}_2\mathrm{O}\) to the right: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \] - Balance hydrogen: Right has 14 H, so add 14 \(\mathrm{H}^+\) to the left: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \] - Balance charge: Left: \(-2 + 14 = +12\); Right: \(2 \times (+3) = +6\). Add 6 electrons to the left: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \] Oxidation half-reaction:
\[ \mathrm{SO}_3^{2-} \rightarrow \mathrm{SO}_4^{2-} \] - Balance oxygen: Left has 3 O, right has 4 O, so add 1 \(\mathrm{H}_2\mathrm{O}\) to the left: \[ \mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} \] - Balance hydrogen: Left has 2 H, so add 2 \(\mathrm{H}^+\) to the right: \[ \mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ \] - Balance charge: Left: \(-2\); Right: \(-2 + 2 = 0\). Add 2 electrons to the right: \[ \mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^- \] -
Step 2: Combine half-reactions
Equalize electrons by multiplying the oxidation half-reaction by 3 (since reduction needs 6 electrons, and oxidation gives 2 electrons per reaction): \[ 3 \left( \mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^- \right) \] \[ 3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ + 6e^- \] Add to the reduction half-reaction: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- + 3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ \] - Electrons cancel out.
- Hydrogen: Left: \(14 \mathrm{H}^+\); Right: \(6 \mathrm{H}^+\). Net: \(14 - 6 = 8 \mathrm{H}^+\) on left. - Water: Left: \(3 \mathrm{H}_2\mathrm{O}\); Right: \(7 \mathrm{H}_2\mathrm{O}\). Net: \(7 - 3 = 4 \mathrm{H}_2\mathrm{O}\) on right.
- Simplify: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 3\mathrm{SO}_3^{2-} + 8\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} \] -
Step 3: Verify balance
- Chromium: 2 left, 2 right.
- Sulfur: 3 left, 3 right.
- Oxygen: Left: \(7 + 9 = 16\); Right: \(12 + 4 = 16\).
- Hydrogen: Left: 8; Right: 8.
- Charge: Left: \(-2 - 6 + 8 = 0\); Right: \(6 - 6 = 0\).
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Step 4: Find the ratio In the balanced equation: \[ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 3\mathrm{SO}_3^{2-} + 8\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} \] The coefficient of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) is 1, and that of \(\mathrm{SO}_4^{2-}\) is 3. Thus, the ratio of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) to \(\mathrm{SO}_4^{2-}\) is: \[ 1 : 3 \] -
Step 5: Match with options
The ratio \( 1:3 \) corresponds to option (C).