6K 3 part a)

Moderators: Chem_Mod, Chem_Admin

ckilkeary 2G
Posts: 68
Joined: Fri Aug 09, 2019 12:16 am

6K 3 part a)

Postby ckilkeary 2G » Sun Mar 01, 2020 10:03 pm

The part of this I'm getting stuck on is the oxidation half-reaction. Could someone explain how they got the oxidation half-reaction step by step of
Cl2 + S2O3 ^2- --> Cl^- + SO4

Jonathan Gong 2H
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

Re: 6K 3 part a)

Postby Jonathan Gong 2H » Sun Mar 01, 2020 10:10 pm

The part of this I'm getting stuck on is the oxidation half-reaction. Could someone explain how they got the oxidation half-reaction step by step of
Cl2 + S2O3 ^2- --> Cl^- + SO4

1) S2O3^2- --> SO4^2- Balance S
2) S2O3^2- --> 2SO4^2- Balance O, Acidic Solution
3) S2O3^2- + 5H2O --> 2SO4^2- Balance H
4) S2O3^2- + 5H2O --> 2SO4^2- + 10H+ Balance electrons
5) S2O3^2- + 5H2O --> 2SO4^2- + 10H+ + 8e-

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: 6K 3 part a)

Postby Justin Vayakone 1C » Sun Mar 01, 2020 10:15 pm

Here is how to balance the oxidation reaction in acidic solution:
Attachments
IMG_3611.JPG

305421980
Posts: 101
Joined: Sat Sep 07, 2019 12:16 am

Re: 6K 3 part a)

Postby 305421980 » Sun Mar 01, 2020 10:17 pm

1) the skeletal half reaction for oxidation is (S2O3)2- --> (SO4)2-
2) balance every element other than O and H, so in this case you would balance S giving you (S2O3)2- --> 2(SO4)2-
3)next you balance O and must keep in mind it takes place in an acidic solution. TO do so, you add H2O to balance giving you 5H2O + (S2O3)2- -> 2(SO4)2-. (eight oxygens on each side)
4)then you add H+ do balance H giving you 5H2O + (S2O3)2- -> 2(SO4)2- + 10H+
5) Finally, you balance the overall charge of each side. The reactants have an overall charge of -2 and the products have +6. Since it is an oxidation reaction you add electrons to the right side. This gives you 5H2O + (S2O3)2- -> 2(SO4)2- + 10H+ + 8e-

Prasanna Padmanabham 4I
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

Re: 6K 3 part a)

Postby Prasanna Padmanabham 4I » Sun Mar 01, 2020 10:25 pm

Here is a prior post I found on Chem Community that goes into every single detail: https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=11377 that solves this question step by step.

But in short, you would 1) write out the Cl and the S2O3 half-reactions separately to first figure out the change in charges, 2) balance out the half-reaction with H+s and H2Os 3) balance charges, 4) put it all together.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 0 guests