Exercise 14.31


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Rucha Kulkarni 2A
Posts: 43
Joined: Fri Sep 29, 2017 7:05 am

Exercise 14.31

Postby Rucha Kulkarni 2A » Sun Feb 18, 2018 7:48 pm

Does anyone know how to solve exercise 14.31?

Mika Sonnleitner 1A
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
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Re: Exercise 14.31

Postby Mika Sonnleitner 1A » Sun Feb 18, 2018 8:37 pm

Step 1: write the half-reactions
Oxidation: Cl2(g) + 2e- --> 2 Cl-(aq)
Reduction: 2Br-(aq) --> Br2(l) + 2e-

Step 2: calculate the standard cell potential for each reaction by looking at Appendix 2B
*you should reverse the anode reaction in the table to get the correct cell potential value for the half-reaction you wrote, which in this case will be -1.09V.
E(cathode)=1.36V

Step 3: calculate standard cell potential for the reaction
E°(cell)=+1.36V-1.09V=+0.27V

Step 4
since E°(cell) is positive, if you plug in a positive value into the equation ΔG°=-nFE°, you would get a negative value for ΔG°. This means the reaction is spontaneous, and therefore favors products.

You can use this same method for the other parts of this problem.

Marina Georgies 1C
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

Re: Exercise 14.31

Postby Marina Georgies 1C » Sun Feb 18, 2018 11:16 pm

to add onto the previous reply, if you don't want to think of it solely in terms of spontaneity, you could also infer that K>1 because Ecell>0, which means products are favored, and this is due to the relationship between E and K, as seen in the equation derived in section 4.8


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