Complete and balance the following redox reaction in basic solution.

Question

Complete and balance the following redox reaction in a basic solution. Be sure to include the proper phases for all species within the reaction.

TeO₃²⁻(aq) + As(s) → Te(s) + H₂AsO₄⁻(aq)

Answer 1

Hopefully by now you would learn how to do these. Follow the steps outlined. They are the same for all of these types of problems, the only difference being if you use H⁺ (acid) or OH^- (base).

TeO₃²⁻(aq) + As(s) → Te(s) + H₂AsO₄⁻(aq) .. unbalanced equation

TeO₃^2- ==> Te .. reduction half reaction (Te goes from +4 to zero)

TeO₃^2- ==> Te + 3H₂O .. balanced for Te and O

TeO₃^2- + 6H₂O ==> Te + 3H₂O + 6OH^- .. balanced for Te, O and H (using base, OH-)

TeO₃^2- + 6H₂O + 4e- ==> Te + 3H₂O + 6OH^- .. balanced for Te, O, H and charge = balanced half reaction

As(s) ==> H₂AsO₄^- .. oxidation half reaction (As goes from zero to +5)

As(s) + 4H₂O ==> H₂AsO₄^- .. balanced for As and O

As(s) + 4H₂O + 6OH^- ==> H₂AsO₄^- + 6H₂O .. balanced for As, O and H (using base, OH-)

As(s) + 4H₂O + 6OH^- ==> H₂AsO₄^- + 6H₂O + 5e- .. balanced for A, O, H and charge = balance half rxn

Multiply reduction rxn by 5 and oxidation rxn by 4 to equalize electrons. Add them and combine/cancel like terms to get the final balanced equation:

5TeO₃^2- + 30H₂O + 20e- ==> 5Te + 15H₂O + 30 OH^-

4As(s) + 16H₂O + 24OH^- ==> 4H₂AsO₄^- + 24H₂O + 20e-

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5TeO₃^2- + 7H₂O + 4As(s) ==> 5Te + 6OH^- + 4H₂AsO₄^- .. balanced redox equation

Answer 2

Using the Oxidation State Method:

TeO₃^2- → Te (Te balanced, now need electrons for Te to go from +4 to 0 in Ox. No.)

4e^- + TeO₃^2- → Te (This is reduction eqn. Now balance with charged species, OH^-)

4e^- + TeO₃^2- → Te + 6OH^- (Now balance O with H₂O and use H balance to check)

3H₂O + 4e^- + TeO₃^2- → Te + 6OH^- (All set)

The oxidation 1/2 rxn:

As → H₂AsO₄ (As is balanced. Ox No goes from 0 to +6, provide electrons)

As → H₂AsO₄ + 6e^-

6OH^- + As → H₂AsO₄ + 6e^-

6OH^- + As → H₂AsO₄ + 6e^- + 2H₂O (All set)

Now just combine the two equations in order to cancel the e^- and leave the net OH^- and H₂O on one side.

Please consider a tutor. Take care.