How do I balance this using half reactions?
(a) Reaction of thiosulfate ion with chlorine gas:
Cl2(g) + S2O32
(aq) -> Cl2(aq) 1 SO422 (aq)
6K3
Moderators: Chem_Mod, Chem_Admin
Re: 6K3
e start off with Cl2 + S2O3 2- --> Cl- + SO4 2-
1) Write half reactions.
Use the Cl's to make a half reaction and the S2O3 to make the other
2) Balance the reaction but disregard O and H (we'll balance those later)
So balance the Cl and the S and we get
S2O3 2- --> 2SO4 2- (we'll call this the 1st HR)
Cl--> 2Cl- for our first half reaction (we'll call this the 2nd HR)
3) Add H2O or H+ to balance the H's and O's
1st HR:
5H2O + S2O3 2- --> 2SO4 2- + 10 H+ . The reason we had a 5 H2O is because we see that S2O3 has 3 O's but it needs to balance with the 8 O's on the products side. Therefore, we add an 5 H2O so each H2O contributes an oxygen to make it 8 oxygens on the reactant side and 8 on the product side.
2nd HR:
Cl2 --> 2Cl- there's no need to add an H+ or H2O as it is balanced. Only the charges need to be balanced
4) Balance the charges
1st HR:
5H2O + S2O3 2- --> 2SO4 2- + 10 H+.
0 -2 --> -4 +10 I wrote down the charges of each reactant/product
The reactants: -2 The products: +6 So how do we go from a -2 charge to a +6 charge. You must lose 8 electrons
5H2O + S2O3 2- --> 2SO4 2- + 10 H+ + 8e- (Remember losing electrons always go on the product side)
2nd HR:
Cl2--> 2Cl-
0 --> -2 So to get from 0 to -2 we need to 2 electrons on the reactant side
2e- + Cl2 --> 2Cl-
5) Cancel out the electrons
we see that HR 1 has 8 electrons and HR 2 has 2 electrons. We have to match HR 2 to have 8 electrons so we multiply the entire HR2 by 4.
2e- + Cl2 --> 2Cl- turns in 8e- +4Cl2-->8Cl-
6) combine the rxns
8e- +4Cl2-->8Cl-
+5H2O + S2O3 2- --> 2SO4 2- + 10 H+ + 8e- You can see the 8e- cancels b/c they are on opposite sides
________________________________________
4 Cl2 + S2O3 2- +5 H2O--> 8 Cl- +2SO4 2- + 10 H+
1) Write half reactions.
Use the Cl's to make a half reaction and the S2O3 to make the other
2) Balance the reaction but disregard O and H (we'll balance those later)
So balance the Cl and the S and we get
S2O3 2- --> 2SO4 2- (we'll call this the 1st HR)
Cl--> 2Cl- for our first half reaction (we'll call this the 2nd HR)
3) Add H2O or H+ to balance the H's and O's
1st HR:
5H2O + S2O3 2- --> 2SO4 2- + 10 H+ . The reason we had a 5 H2O is because we see that S2O3 has 3 O's but it needs to balance with the 8 O's on the products side. Therefore, we add an 5 H2O so each H2O contributes an oxygen to make it 8 oxygens on the reactant side and 8 on the product side.
2nd HR:
Cl2 --> 2Cl- there's no need to add an H+ or H2O as it is balanced. Only the charges need to be balanced
4) Balance the charges
1st HR:
5H2O + S2O3 2- --> 2SO4 2- + 10 H+.
0 -2 --> -4 +10 I wrote down the charges of each reactant/product
The reactants: -2 The products: +6 So how do we go from a -2 charge to a +6 charge. You must lose 8 electrons
5H2O + S2O3 2- --> 2SO4 2- + 10 H+ + 8e- (Remember losing electrons always go on the product side)
2nd HR:
Cl2--> 2Cl-
0 --> -2 So to get from 0 to -2 we need to 2 electrons on the reactant side
2e- + Cl2 --> 2Cl-
5) Cancel out the electrons
we see that HR 1 has 8 electrons and HR 2 has 2 electrons. We have to match HR 2 to have 8 electrons so we multiply the entire HR2 by 4.
2e- + Cl2 --> 2Cl- turns in 8e- +4Cl2-->8Cl-
6) combine the rxns
8e- +4Cl2-->8Cl-
+5H2O + S2O3 2- --> 2SO4 2- + 10 H+ + 8e- You can see the 8e- cancels b/c they are on opposite sides
________________________________________
4 Cl2 + S2O3 2- +5 H2O--> 8 Cl- +2SO4 2- + 10 H+
-
- Posts: 98
- Joined: Wed Sep 18, 2019 12:18 am
Re: 6K3
Why is Cl2 the oxidizing agent when it actually gains electrons? Shouldn't it be the reducing agent in that case?
-
- Posts: 111
- Joined: Wed Sep 18, 2019 12:18 am
Re: 6K3
It is the oxidizing agent because it is the one that causes oxidation to happen. It ends up losing electrons by causing another one to gain electrons I think is my understanding of it
Return to “Kinetics vs. Thermodynamics Controlling a Reaction”
Who is online
Users browsing this forum: No registered users and 4 guests