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Solution:The equation is:Oxidation half-reaction:
S + 2H2O ---> SO2 +4H+ + 4e- Reduction half-reaction: 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 +4H+ ---> 3SO2 + 4NO + 2H2O
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