## Midterm 2013, Q7

ribal_daniel_3H
Posts: 13
Joined: Wed Sep 21, 2016 2:58 pm

### Midterm 2013, Q7

Q7 "Calculate K and dG at 298K for the following redox reaction. Hint: Identify the two half-reactions and show a balanced equation to receive full credit
Mn^2+aq + Br2(l) ----> MnO4^-aq + Br^-(aq)

The question gives us the half reactions for MnO4 and Br2 and adds them up, but flips the half reaction for MnO4. How come MnO4 is flipped?

Thomas_Elizabeth_3A
Posts: 11
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Midterm 2013, Q7

Instead of having to deal with flipping the reaction, you could use the method of Ecell=Ecathode-Eanode. With this method you just figure out which half reaction is the anode and which is the cathode. In this problem the Br reaction is the cathode and the MnO4 is the anode, so then you subtract the E values for each (1.07 for Br reaction - 1.49 for MnO4 reaction) and get -0.42 for the Ecell.

ChristineCastanon_1B
Posts: 10
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Midterm 2013, Q7

MnO4 is flipped because it is the half-reduction that is being oxidized. MnO4 is loosing electrons during the reaction, making it the anode and the anode is always negative.

Alexxa_Vasquez_3B
Posts: 11
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Midterm 2013, Q7

^ And with that logic, since we know that the cathode (+) is gaining electrons and the anode is giving electrons, that can help us remember that the cathode is the reduction half reaction and the anode is the oxidation half reaction. :]

ribal_daniel_3H
Posts: 13
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Midterm 2013, Q7

Okay, I think I understand now. With all of the logic provided, I think I will be able to figure it out during the midterm. Thank you all :]

Fayt Sarreal 1G
Posts: 25
Joined: Fri Jul 22, 2016 3:00 am

### Re: Midterm 2013, Q7

Anyone here know if we will need to balance out the hydrogen and oxygen the half reactions for the midterm?