Consider the following unbalanced redox equation: H+ + Cr2O72– + CH3CHO → CH3COOH + Cr3+ + H2O When this equation is completely balanced using simplest whole number ratio, the number of protons required will be

Question

Consider the following unbalanced redox equation: H+ +  Cr2O72– +  CH3CHO →  CH3COOH +  Cr3+ +  H2O When this equation is completely balanced using simplest whole number ratio, the number of protons required will be

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To balance this redox equation in acidic solution, I will separately balance the oxidation and reduction half-reactions, then combine them.
First, identify the oxidation half-reaction: CH3CHO (acetaldehyde) → CH3COOH (acetic acid). In acidic media, aldehydes oxidize to carboxylic acids by releasing protons and electrons. A common balanced form is: CH3CHO + H2O → CH3COOH + 2H+ + 2e-. This shows that the oxidation half-reaction produces 2 electrons per molecule oxidized.
Next, balance the reduction half-reaction for Cr2O7^2− to Cr3+. The standard acidic reduction half-reaction is: Cr2O7^2− + 14H+ + 6e− → 2Cr3+ + 7H2O. This consumes 6 electrons per dichromate ion.
To balance the overall reaction, the electrons must cancel. The oxidation half-reaction produces 2e− per CH3CHO, so multiply it by 3 to yield 6e− total: 3CH3CHO + 3H2O → 3CH3COOH + 6H+ + 6e−.
Now add the two half-reactions together:
Cr2O7^2− + 14H+ + 6e− → 2Cr3+ + 7H2O
+ (3CH3CHO + 3H2O → 3CH3COOH + 6H+ + 6e−)
Cancel the electrons on both sides (6e−). Combine the species: the H+ on the left is 14H+, on the right there are 6H+ added from the oxidation, so net H+ on the left becomes 14H+ − 6H+ = 8H+. Water counts: there are 3H2O on the oxidation side and 7H2O on the chromium reduction side; moving terms shows a net 4H2O on the product side.
Putting it all together, the balanced equation is:
Cr2O7^2− + 8H+ + 3CH3CHO → 2Cr3+ + 4H2O + 3CH3COOH.
From the balancing steps, the number of protons required on the reactant side is 8.
Now evaluate the answer options:
- Option a: 8 — This matches the calculated proton count, so it is a plausible correct choice.
- Option b: 6 — This would correspond to fewer protons than balanced here; it would not cancel the electrons properly with the dichromate reduction. It indicates a miscount of H+ in the overall sum.
- Option c: 14 — This equals the standard Cr2O7^2− reduction proton count alone, but after combining with the oxidation half-reaction, some protons cancel or are consumed/produced, yielding a smaller net number. Selecting 14 would disregard the 6 protons produced by oxidation being canceled in the overall process.
- Option d: 2 — This is far too low and would not balance the chromium and oxygen/hydrogen counts; it also ignores the protons produced by oxidation and the stoichiometry of the reduction half-reaction.
In summary, only 8 protons are required in the completely balanced equation, while the other options reflect common miscounts or misunderstandings of halving, canceling, or balancing proton terms across half-reactions.

MUF0042 Chemistry Unit 2 - Semester 2, 2025 10.13 Quiz: Balancing Redox Reactions

Consider the following unbalanced redox equation: H+ + Cr2O72– + CH3CHO → CH3COOH + Cr3+ + H2O When this equation is completely balanced using simplest whole number ratio, the number of protons required will be

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