The quantity of electricity in Faraday needed to reduce 1 mol of Cr₂O₇²⁻ to Cr³⁺ is ____________.

Solution

Chemistry ma’am · 4 steps

Chromium goes from +6 to +3, and there are two Cr atoms, so count the electrons twice over.

1. Oxidation state of Cr. In Cr₂O₇²⁻, let Cr be x. Then 2x + 7(-2) = -2, so 2x = 12 and x = +6.
2. Electrons per Cr. Each Cr goes from +6 to +3, so each one gains 3 electrons. One dichromate ion has two Cr atoms, so it takes 2 × 3 = 6 electrons.

   Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

3. Electrons to Faraday. 1 mol of electrons carries 1 Faraday of charge. So 1 mol of Cr₂O₇²⁻ needs 6 mol of electrons, that is 6 F.
4. So. The quantity of electricity needed is 6 Faraday.

The common slip: 3. Counting only one Cr atom (+6 to +3) gives 3, not 6. Dichromate has two Cr atoms, so double it.

The idea to keep. Faradays needed equal moles of electrons transferred. Find the change in oxidation state per atom, then multiply by how many of those atoms one formula unit holds.

Answer from the NTA final answer key. Solution by PrepWiser.