2so2 O2 2so3 Redox Reaction
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How exercise I balance this reaction using half-redox equations? $$\ce{2SO2 + O2->2SO3}$$
My work is:
Oxidation: $\ce{SO2 + Water -> SO3 + 2e- + 2H+}$
Reduction: $\ce{H2o + 2e- + O2 -> SO3 + 2H+}$
Total: $\ce{SO2 + 2H2O + O2 -> 2SO3 + 4H+}$
Merely it's false. What did I do wrong?
tschoppi
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asked Nov 10, 2014 at 12:45
user9686user9686
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ane
1 Answer 1
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WELL I think you put something erroneous in your reduction formula.
You volition never get SO3 out of simply O2 and water. Instead of SO3
The reduction formla will be 4e- + O2 + 4H+ --> 2 Water
Multiply oxidation formula past 2 and add it to new reduction formula and you got your answer
answered November 10, 2014 at 16:21
DSinghviDSinghvi
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2so2 O2 2so3 Redox Reaction,
Source: https://chemistry.stackexchange.com/questions/19355/how-can-i-balance-the-redox-reaction-of-sulfur-dioxide-with-oxygen
Posted by: riveragraters.blogspot.com
$\begingroup$ Your reduction formula is wrong. For i thing you have no sulfur containing species on the reactant side and the reactant side has a charge of -2 while the product side shows a charge of +ii, this is a violation of accuse neutrality, both sides must have the same overall charge. $\endgroup$
Nov 10, 2014 at 15:32