Use oxidation states to identify the element that is oxidized and the element that is reduced in the redox reaction. S(s)+6HNO3(aq)→H2SO4(aq)+6NO2(g)+2H2O(l)
Oxygen and hydrogen both follow the rules associated with
oxidation numbers as neither of them are an exception in this
Hydrogen is always +1, except when it is bonded to a group 1 metal (alkali metal)
Oxygen is always -2, expect when it is a peroxide.
Hence in the above reactions the only atoms that are present other than H and O are S and N we will be considering them to estimate which is oxidised and which is reduced.
So to balance the charge sulphur must have the remaining charge The sulfur must balance this remaining negative charge.S = +6.
So sulphur is oxidised from 0 state to +6 state.
In HNO3 H is +1and O is 3x -2 = -6 so N is +5.
In NO2 N is +4 When N is converted from HNO3 to NO2 it oxidation state changes from +5 to +4 so it is reduced. So 6 nitrogen get reduced by 1 each and in turn 1 sulphur atom gets oxidies by +6.
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