14.3 Balancing redox reactions

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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.


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