What is the balanced chemical equation for each half-reaction and their overall reaction?

Question

Write out the balanced chemical equation for each half-reaction and their overall reaction?

Anode: Cr(s) → Cr⁺³(aq)

Cathode: I₂(aq) → I^-1(aq)

Anode: NO₃^-1aq) → HNO₂ (aq)

Cathode: Au⁺³(aq) → Au(s)

Anode: ClO₄^-1(aq) → Cl₂(aq)

Cathode: Cr₂O₇^-2(aq) → Cr⁺³(aq)

Answer 1

Oxidation takes place at the anode. Reduction takes place at the cathode.

oxidation: Cr(s) ==> Cr³⁺(aq) + 3e-

reduction: I₂(aq) + 2e- ==> 2I^-(aq)

Overall: 2Cr(s) + 3I₂ ==> 2Cr³⁺ + 6I^-

NO₃^- ==> HNO₂ this is a reduction reaction (N goes from +5 to +3)

Au³⁺ => Au this is a reduction reaction (Au goes from +3 to zero)

You can't have 2 reduction reactions and still have a galvanic cell

If they want you to reverse the anode reaction to an oxidation, then you can write the overall reaction

oxidation: 3HNO₂ + 3H₂O ==> 3NO₃^- + 9H⁺

reduction: 2Au³⁺ ==> 2Au

overall: 3HNO₂ + 2Au³⁺ + 3H₂O ==> 3NO₃^- + 2Au + 9H⁺

2ClO₄^- ==> Cl₂ this is a reduction (Cl goes from +7 to zero)

Cr₂O₇^2- ==> Cr³⁺ this is a reduction (Cr goes from +6 to +3)

You can't have 2 reduction reactions and still have a galvanic cell

If they want you to reverse the anode reaction to an oxidation, then you can write the overall reaction (see b)