Anode and cathode

Moderators: Chem_Mod, Chem_Admin

904638389
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Anode and cathode

Postby 904638389 » Wed Feb 10, 2016 11:49 am

When I am given two half reaction and the electrons are both on the left side (because E is given that way) and both values of E are given. How do I tel l which one is the anode and cathode?

Alisha Clark
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

Re: Anode and cathode

Postby Alisha Clark » Wed Feb 10, 2016 1:49 pm

Which ever combination makes the overall charge of the cell more positive.

Mikyla Reta 2J
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am

Re: Anode and cathode

Postby Mikyla Reta 2J » Wed Feb 10, 2016 2:10 pm

UA Lauren said the anode is always the smallest E naught value.
So if one value is positive and one is neg, its the neg
If both are pos its the smallest pos value
if both are neg its the larger neg value (making it the smaller than the other value)

Ryan McDonough 4E
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: Anode and cathode

Postby Ryan McDonough 4E » Thu Feb 11, 2016 4:01 am

However, note that every so often, there are exceptions to this rule. Take question 8 from the Winter 2014 Midterm Practice Test.

We are given two half reactions and asked to calculate the Ka for HF. To find this, we want to make HF a reactant. By flipping the equation, we make the HF equation the anode, and the E naught cell potential is -0.16 which is given when we subtract the cathode E naught (2.87) by the anode E naught (3.03). By doing this the equation looks like this
ln([H]^2[F]^2 / [HF]^2) = (n*F*E naught)/(RT).

By breaking the rule of finding the largest E naught cell potential, we are able to find the Ka value of HF. Exceptions to the rule are rarer, but they are there. Hope this helps :).


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest