HW 14.3 Balancing Half Reactions in Acidic Solutions

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Aimiel Casillan 1J
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Joined: Fri Sep 25, 2015 3:00 am

HW 14.3 Balancing Half Reactions in Acidic Solutions

Postby Aimiel Casillan 1J » Thu Feb 04, 2016 1:07 am

I'm having trouble balancing the half reactions of acidic solutions, such as those in 14.3 from the Electrochem HW, especially when to use H+ or extra electrons in the equation. Using any of the subproblems in 14.3, could someone please outline the steps from identifying the oxidizing/reducing agents to balancing the half redox reactions in order to achieve the overall redox reaction?

Thanks so much!

Joshua Kim 3H
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: HW 14.3 Balancing Half Reactions in Acidic Solutions

Postby Joshua Kim 3H » Thu Feb 04, 2016 1:39 am

Let's look at 14.3 a
We 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+ Final answer
Make sure to add the correct states too i.e (aq) or (g).


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