Test 2 Question 6 Part B

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

Test 2 Question 6 Part B

Postby Gwen Peng 1L » Sat Mar 17, 2018 7:06 pm

For this question the galvanic cell is formed with 03/02, OH- and )3, H+/O2 and it gave the following half reactions:
O3 + H2O + 2e- --> O2 + 2OH-
O3 + 2H+ + 2e- --> O2 + H2O
If the oxidizing agent (cathode) is O3 then what is the reducing agent (anode)?

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

Re: Test 2 Question 6 Part B

Postby Justin Bui 2L » Sat Mar 17, 2018 7:23 pm

For this problem, the detachable sheet with the reduction potentials is handy to have because it will tell you V, and that will help you determine which half reaction will be reduction/oxidation. This is because you want your Ecell to be positive, so you would want Ecathode to be more positive than Eanode to make sure Ecell is positive. After that's established, you can reverse the appropriate equation to complete the full net equation to find that O2 is the reducing agent.

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