Consider the following ionic equation: Cr$_2$O$_7$$^{2-}$ + 14H$^+$ + ne-  → 2Cr$^{3+}$ + 7H$_2$O. The value of n in the equation is_______?

Analyze the question: We need the number of electrons (n) that balances the half-equation, which shows dichromate being reduced to chromium(III).

Concept: In a balanced ionic equation, both mass and charge must balance. The total charge on the left must equal the total charge on the right. You can also use oxidation numbers: the electrons gained equal the total drop in oxidation number.

Solution (Method 1: charge balance)

Left side:

• Cr₂O₇²⁻ has a charge of −2
• 14H⁺ has a charge of +14
• n electrons have a charge of −n

Total charge on the left = −2 + 14 − n = 12 − n

Right side:

• 2Cr³⁺ has a charge of +6
• 7H₂O has a charge of 0

Total charge on the right = +6

Equate them: 12 − n = 6, so n = 6.

Solution (Method 2: oxidation numbers)

In Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = 12 and x = +6 for each Cr. In Cr³⁺: the oxidation number is +3.

Each Cr gains 3 electrons (+6 → +3). There are 2 Cr atoms, so 2 × 3 = 6 electrons.

Both methods agree: n = 6.

Answer: B. 6.

Tip: For redox half-equations, always check the charge on both sides. Also remember that the dichromate ion, Cr₂O₇²⁻, is an oxidizing agent in acid, where Cr(VI) turns from orange to green Cr(III) by gaining 6 electrons in total (3 per Cr atom).