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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-?

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dsdrajlin

Answer:

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.

  • 1 mole of Pb is oxidized when 2 moles of electrons circulate.
  • 1 mole of Cr₂O₇²⁻ is reduced when 6 moles of electrons circulate.

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

[tex]3molCr_2O_7^{2-} \times \frac{6mol\ e^{-} }{1molCr_2O_7^{2-}} \times \frac{1molPb}{2 mol\ e^{-} } = 9molPb[/tex]

pstnonsonjoku

According to the information in the question, 9 moles of Pb(s) is oxidized by three moles of dichromate ion.

A redox reaction involves loss and gain of electrons. The oxidizing agent gains electrons while the reducing agent looses electrons. For the reaction stated in the question;

Oxidation half equation;

6Pb(s) → 6Pb2+(aq) + 12 e-

Reduction half equation;

2Cr2O72-(aq) + 28 H+(aq) + 12 e- → 4 Cr3+(aq) + 14 H2O(l)

The overall equation is;

6Pb(s) + 2Cr2O72-(aq) + 28 H+(aq)  → 6Pb2+(aq) + 4 Cr3+(aq) + 14 H2O(l)

If 6 moles of  Pb(s) is oxidized by 2 moles of Cr2O72-

x moles of  Pb(s) is oxidized by 3 moles of Cr2O72-

x =  6 moles × 3 moles/ 2 moles

x = 9 moles of Pb(s)

Learn more: https://brainly.com/question/8646601

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