To find the ratio of to 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: - Oxidation: Reduction half-reaction:
- Balance chromium: Left has 2 Cr, right has 2 Cr. - Balance oxygen: Left has 7 O, so add 7 to the right: - Balance hydrogen: Right has 14 H, so add 14 to the left: - Balance charge: Left: ; Right: . Add 6 electrons to the left: Oxidation half-reaction:
- Balance oxygen: Left has 3 O, right has 4 O, so add 1 to the left: - Balance hydrogen: Left has 2 H, so add 2 to the right: - Balance charge: Left: ; Right: . Add 2 electrons to the right: -
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): Add to the reduction half-reaction: - Electrons cancel out.
- Hydrogen: Left: ; Right: . Net: on left. - Water: Left: ; Right: . Net: on right.
- Simplify: -
Step 3: Verify balance
- Chromium: 2 left, 2 right.
- Sulfur: 3 left, 3 right.
- Oxygen: Left: ; Right: .
- Hydrogen: Left: 8; Right: 8.
- Charge: Left: ; Right: .
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Step 4: Find the ratio In the balanced equation: The coefficient of is 1, and that of is 3. Thus, the ratio of to is: -
Step 5: Match with options
The ratio corresponds to option (C).