6th edition, 14.3b

Moderators: Chem_Mod, Chem_Admin

Hai-Lin Yeh 1J
Posts: 89
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

6th edition, 14.3b

Postby Hai-Lin Yeh 1J » Wed Feb 27, 2019 8:14 pm

Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction.
(a) Reaction of thiosulfate ion with chlorine gas:
Cl2 (g) + S2O3 2- (aq) -> Cl- (aq) + SO4 2- (aq)
I am trying to balance the oxidation half-reaction, and so far I have:
5H2O +S2O3 2- -> 2SO4 2- + 10H+
Why do we add 8e- not 4e- since the left has a charge of 2- and the right as a charge of 4- + 10+ which is 6+, so shouldn't we add 4 e- to match the left?

chaggard
Posts: 37
Joined: Fri Sep 28, 2018 12:19 am

Re: 6th edition, 14.3b

Postby chaggard » Wed Feb 27, 2019 8:23 pm

So the reaction of S2O3^2- --> SO4^2- is a oxidation reaction because the Chlorine reaction is reduction. Starting off, the left side of the reaction we calculate 2 Sulfur (S) molecules to have a charge of +4 because x-6=-4 so x=+4. On the right, we add a coefficient of 2 in front to make the sides balanced. This means that when we solve for x (oxidation number), you have to remember to multiply the x value by 2 because there are 2 SO4^2- molecules. So, x-8=-2, so x=-6, and multiply by 2 we get 2x=+12. The left is +4, the right is +12, so 8e- need to be added to the right side. Hope this helps!


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest