6K.5 part a

Moderators: Chem_Mod, Chem_Admin

VioletKo3F
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

6K.5 part a

Postby VioletKo3F » Thu Mar 05, 2020 3:17 pm

How are we supposed to balance out H2O? Can someone walk me through the steps/reasoning?

William Francis 2E
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

Re: 6K.5 part a

Postby William Francis 2E » Thu Mar 05, 2020 5:22 pm

I'm a bit confused about what you mean. I don't see H2O in the skeletal equation given, but it does need to be added to balance the equation. The first skeletal half reaction is Br-(aq)-->BrO3-(aq). The oxidation number of Br increases from -1 to +5 in this, so we know it is the oxidation half reaction. Firstly, 3H2O(l) must be added to the left side of the equation to balance oxygen. Then, 6H2O(l) should be added to the right side, and 6OH- should be added to the left to balance hydrogen (this process is described in toolbox 6k.1 in the textbook). Then, you add electrons to balance charges on each side of the half reaction and cancel out 3H2O(l) from each side since each side has that. This will leave 3H2O(l) on the right and no water molecules on the left side. The same process can be done to balance the other skeletal half reaction.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest