Question

Because molecular bromine is a string oxidant (and super toxic), it is necessary to "neutralize" any remaining bromine at the end of the reaction period. That is the reason for the addition of the sodium thiosulfate solution. Sodium thiosulfate is a good reducing agent and converts molecular bromine to (the innocuous) sodium bromide. The thiosulfate anion is converted to sulfate ion in the process. Omitting the sodium spectator ions, write and balance the oxidation and reduction half-reactions that are occuring here. (Remember you may use H+/H2O to attain atom balance) Then answer the questions that follow. The oxidation number of sulfur changes from ______ (thiosulfate anion) to _______ (sulfate anion) The oxidation number of bromine changes from _____ (Br2) to _____ (bromide anion).

Answer 1

Reaction : S2O3^2- + Br2 ---> SO4^2- + Br-

S2O3^2- ---> SO4^2-

balance sulfur and Oxygen,

S2O3^2- + 5H2O ---> 2SO4^2-

balance H and balance e-,

S2O3^2- + 5H2O ---> 2SO4^2- + 10H+ + 4e-

And,

Br2 ---> Br-

balance Br and e-

Br2 + 2e- ---> 2Br-

multiply with 2

2Br2 + 4e- ---> 4Br-

Add both equations,

S2O3^2- + 2Br2 + 5H2O ----> 2SO4^2- + 4Br- + 10H+

is the balanced equation

The oxidation number of sulfur chages from +2 (thiosulfate anion) to +6 (sulfate anion). The oxidation number of bromine changes from 0 (Br2) to -1 (bromide anion).