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Solution:

The equation is:
S + NO3- ---> SO2 + NO

Oxidation half-reaction:

S ---> SO2
S + 2H2O ---> SO2 +4H+ + 4e-

Reduction half-reaction:

NO3- ---> NO
NO3_ + 4H+ + 3e- ---> NO+2H2O

To have the same number of electrons in both half-reactions, multiply the oxidation reaction by 3 and multiply the reduction reaction by 4. After this, add the two half-reactions together and then cancel:

4NO3- +3S +16H+ +6H2O ---> 3SO2 + 16H+ + 4NO +8H2O

4NO3- + 3S +4H+ ---> 3SO2 + 4NO + 2H2

Back to the Redox Page

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