## Exercise 14.31

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

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

### Exercise 14.31

Does anyone know how to solve exercise 14.31?

Mika Sonnleitner 1A
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### Re: Exercise 14.31

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

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