Eletro chem Test

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Juanalv326
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

Eletro chem Test

Postby Juanalv326 » Sat Mar 17, 2018 12:24 pm

How do you know how to solve for the reducing agent and oxizing agent for the reaction, O3/O2,OH and O3, H+/O2

Gevork 2E
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am
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Re: Eletro chem Test

Postby Gevork 2E » Sat Mar 17, 2018 12:29 pm

Hi,

You have to see which molecule in the pair is being reduced/oxidized and then go from there. I don't entirely understand the example you wrote so sorry if the question was specific to that example. If you are going from O2 to OH-, then your oxygen goes from a charge of 0 to a charge of -2. It is gaining electrons, thus it is being reduced and therefore it is the oxidizing agent.

Hope this was helpful.

Cynthia Bui 2H
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: Eletro chem Test

Postby Cynthia Bui 2H » Sat Mar 17, 2018 5:35 pm

I'm not quite sure if this is what you wanted but if for the second one its from H3O+ to O2, then the O is going from 1+ to 0, so it would be reduced and the oxidizing agent.

Justin Bui 2L
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

Re: Eletro chem Test

Postby Justin Bui 2L » Sat Mar 17, 2018 7:11 pm

So I think what you need to do is look on the detachable sheet that contains all the reduction potentials and look for the ones that apply to O3/O2,OH- and O3,H+/O2. Then, based on what the reduction potentials are, you can decide which half reaction should be the cathode/anode because you want the Ecell to be positive and knowing the V for each will help you set that up. From there, you just need to look at the charges on the things in each to see what is the agent for what. Oxidizing agent in this case is O3 and reducing agent is O2.

Gwen Peng 1L
Posts: 36
Joined: Sat Jul 22, 2017 3:01 am

Re: Eletro chem Test

Postby Gwen Peng 1L » Sat Mar 17, 2018 11:07 pm

To solve you need the 2 half reactions:
O3 + H2O + 2e- --> O2 + 2OH- E standard= 1.24 V
O3 + 2H+ + 2e- --> O2 + H2O E standard= 2.07 V
Your cathode is always the oxidizing agent since it undergoes reduction and anode is always the reducing agent since it undergoes oxidation.
Your anode is always the half reaction with a higher voltage because it will be flipped changing the sign and making it negative, and when a negative value is plugged into E standard (cell)= E standard (cathode) - E standard (anode) the double negative cancels and gives a positive overall standard cell potential.
Therefore, you know the second half reaction will be flipped and the reducing agent is O2 and the oxidizing agent is O3.
Hope this helps :)


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