14.3 Balancing redox reactions

Moderators: Chem_Mod, Chem_Admin

Vanessa Romero-Campos 2B
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

14.3 Balancing redox reactions

Postby Vanessa Romero-Campos 2B » Fri Feb 16, 2018 9:08 pm

a)Cl2 + S2O3(-2) --->Cl (-1) + SO4 (-2)

I am trying to balance the reaction and so far I have Cl2+2e-->2Cl- and S2O3(-2)+5H2O---> 2SO4 (-2) + 10H+
How do I finish balancing: S2O3(-2) + 5H2O---> 2SO4 (-2) + 10H+

Abigail Yap 2K
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

Re: 14.3 Balancing redox reactions

Postby Abigail Yap 2K » Fri Feb 16, 2018 9:25 pm

The last step is to balance the charges. You currently have a charge of -2 on the left side of the reaction, and a charge of 2(-2) + 10 = +6 on the right side of the reaction. Adding 8 electrons to the products side will balance the charges.

Your final balanced reaction will be:
S2O3(2-) + 5H2O -> 2SO4(2-) + 10H+ + 8e-

Juanyi Tan 2K
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.3 Balancing redox reactions

Postby Juanyi Tan 2K » Sat Feb 17, 2018 2:49 am

I agree with Abigail! The two things we should look for when we balance the redox equation are the charge and the number of elements.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 15 guests