For a galvanic cell that uses the following two half-reactions, Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) Pb(s) → Pb2+(aq) + 2 e- how many moles of Pb(s) are oxidized by three moles of Cr2O72-?

9 mol

Explanation:

Let's consider the following half-reactions taking place in a galvanic cell.

Reduction: Cr₂O₇²⁻(aq) + 14 H⁺(aq) + 6 e⁻ → 2 Cr³⁺(aq) + 7 H₂O(l)

Oxidation: Pb(s) → Pb²⁺(aq) + 2 e⁻

We can establish the following relationships.

The moles of Pb(s) are oxidized by three moles of Cr₂O₇²⁻ are:

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