What is the value of n in the following equation? Cr2O72- + 14H+ + ne- → 2Cr 3+ + 7H2O

Analyze the question: We need the number of electrons (n) that balance the half-equation for the reduction of dichromate ions to chromium(III) ions.

Concept: A balanced equation must have the same atoms and the same total charge on both sides. Since the atoms are already balanced here, we use charge balance to find n.

Solution:

Left side charge:

• Cr₂O₇²⁻ gives −2
• 14H⁺ gives +14
• n electrons give −n

Total left = −2 + 14 − n = 12 − n

Right side charge:

• 2Cr³⁺ gives 2 × (+3) = +6
• 7H₂O gives 0

Total right = +6

Set them equal:

12 − n = 6 n = 6

Check with oxidation states: Each Cr in Cr₂O₇²⁻ is +6 (2x + 7(−2) = −2, so x = +6). Each Cr goes from +6 to +3, gaining 3 electrons per Cr. For 2 Cr atoms, that is 2 × 3 = 6 electrons. This agrees.

Answer: n = 6, which is option C.

Tip: For redox half-equations, always do a quick charge-balance check, with total charge on the left equal to total charge on the right. You can also count the change in oxidation state per atom and multiply by the number of atoms. Cr going from +6 to +3 across 2 atoms means 6 electrons.