Balancing Redox in Acids and Bases

Moderators: Chem_Mod, Chem_Admin

kshalbi
Posts: 68
Joined: Fri Sep 25, 2015 3:00 am

Balancing Redox in Acids and Bases

Postby kshalbi » Tue Feb 02, 2016 10:15 pm

How would I know to balance an equation the acid way or the base way if they don't specify in the question. For example, 14.15 part C says: Cd+2Ni (OH)3---> Cd (OH)2....Do I use the method for bases because there is an OH involved?

Daniella Ching 4C
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

Re: Balancing Redox in Acids and Bases

Postby Daniella Ching 4C » Tue Feb 02, 2016 11:29 pm

Since they give you the entire reaction, Cd(s) + 2 Ni(OH)3(s) -> Cd(OH)2(s) + 2 Ni(OH)2(s)
you can balance the half reactions by first separating Cd and Ni into separate reactions.
The first one would be: Cd(s) -> Cd(OH)2(s)
and Ni would be: 2 Ni(OH)3(s) -> 2 Ni(OH)2(s)

Then you can add OH- accordingly to balance out the reaction. Therefore, the Cd half reaction would be:
2OH- + Cd(s) -> Cd(OH)2(s)
and the Ni half reaction would be:
2 Ni(OH)3(s) -> 2 Ni(OH)2(s) + 2OH-
(If the two sides were unbalanced by just missing hydrogen, then you would just add H+ to balance it. Since its missing both oxygen and hydrogen, then you add OH-).

Then you can balance the electrons for both half reactions, since now the sides with OH- has a -2 charge (so the other sides---products for Cd and reactants for Ni---have 2 electrons added to them to make it an overall neutral charge for the half reaction).

By approaching the problem like this, you don't need to memorize an acid way or a base way (you would just add H+ or OH- accordingly after seeing what makes the half reaction unbalanced). Hopefully this helps!


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 2 guests