How do you work out the oxidation number of sulfur in SO2
Sulfur in H2SO4 has an oxidation number of +6 how do you know the ox number had been reduced from 6+ to 4+? (It's +4 in sulfur dioxide)
H2SO4 + 2 H+ -> SO2 + 2H2O
H2SO4 + 2 H+ -> SO2 + 2H2O
Original post by Nurhayat
Sulfur in H2SO4 has an oxidation number of +6 how do you know the ox number had been reduced from 6+ to 4+? (It's +4 in sulfur dioxide)
H2SO4 + 2 H+ -> SO2 + 2H2O
H2SO4 + 2 H+ -> SO2 + 2H2O
Well net charge of zero so all the ox states balance out.... O has a -2 in all the compounds you'll encounter.
Original post by JMaydom
Well net charge of zero so all the ox states balance out.... O has a -2 in all the compounds you'll encounter.
SO2 = 0
S + -2 = 0
S = +2
How would You work it out because my one is incorrect
Original post by Nurhayat
SO2 = 0
S + -2 = 0
S = +2
How would You work it out because my one is incorrect
S + -2 = 0
S = +2
How would You work it out because my one is incorrect
O2 is -4 (2*-2), therefore your sulphur must be +4 to give an overall of zero.
Original post by Sammydemon
O2 is -4 (2*-2), therefore your sulphur must be +4 to give an overall of zero.
Thanks!
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