You now have two half-reactions in which the elements are balanced (shown below)

Question

You now have two half-reactions in which the elements are balanced (shown below) but before you sum them, you need to account for the transfer of showed you how to make certain the electrons properly cancel when summing the reduction and oxidation half-reactions by making sure the total number electron on the reactant side of the reduction half-reaction and the product side of the oxidation half-reaction were equal. Once you determine the factors you multiply each half-reaction to balance the electrons, sum the half-reactions. Since it is possible for each half-reaction to contain hydroxide ion and molecules, you may need to sum coefficients and/or cancel chemical species depending on the reaction (like you did for water in Parts B and C). The reaction equation should represent the net totals for all chemical at species. 6e^- + 2BrO_2^- (aq) + 4H_2O9l) rightarrow Br_2(l) + 8OH^-(aq) S_2O_2^2- (aq) + 6OH(aq) rightarrow 2SO_3^2 (aq) + 3H_2 O9l) + 4e^- The mass balanced and charge balanced half-reactions that you determined in parts B and C are shown again in the transition above. Write the final chemical equation for the radix reaction based on the two half-reaction (the reduction in part B and condition reaction in part C.) Express your answer as a chemical equation phases.

Explanation / Answer

Answer – We are given the two half-redox reactions and we need to write the overall net redox reaction.

Reactions are as follow –

6e- + 2BrO2-(aq) + 4 H2O(l) ----> Br2(l) + 8 OH-(aq)

         S2O32-(aq) + 6 OH-(aq) -----> 2SO32-(aq) + 3H2O(l) + 4e-

Now we need the balance the electrons from both half reactions. So need to multiply first half reaction with 2 and the second reaction multiply with 3, so reactions formed as follow –

12e- + 4 BrO2-(aq) + 8 H2O(l) ----> 2 Br2(l) + 16 OH-(aq)

+     3 S2O32-(aq) + 18 OH-(aq) -----> 6 SO32-(aq) + 9 H2O(l) + 12e-

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4 BrO2-(aq) + 3 S2O32-(aq) + 2 OH-(aq) ----> 2 Br2(l) + 6 SO32-(aq) + H2O(l)

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