## Test #2 Question 5B

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

Anne 2L
Posts: 35
Joined: Fri Sep 29, 2017 7:05 am

### Test #2 Question 5B

Rank the following in order of increasing reducing power:

I2, Br2, Al

I2 (s) + 2e- -> 2I- $E^{\circ }$ = 0.54V
Br2 (l) +2e- -> 2Br- $E^{\circ}$ = 1.07V
Al3+(aq) + 3e- -> Al(s) $E^{\circ}$ = -1.66V

Correct Answer: Br2 < I2 < Al

I understand that reducing power increases as $E^{\circ}$ becomes increasingly negative, which makes sense with the data given. I was just wondering why we didn't have to reverse the reactions involving I2 and Br2, because I2 and Br2 are both in the product. And if we were to reverse the reaction, would we add a negative to $E^{\circ}$?

Ishan Saha 1L
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

### Re: Test #2 Question 5B

Hi, I2 and Br2 are actually in the reactants side. They accept 2e- and thus become 2I- ad 2Br-. We don't reverse these equations because this is naturally what happens. I think what you were talking about was whether or not we had to flip the equation for Al because that is in the product side. We don't have to do this while ranking elements in order of reducing power because Al doesn't gain electrons and Al3+ gains 3e- which is why the standard reduction equation is written Al3+(aq) + 3e- ==> Al(s). If we needed to reverse the reaction to solve a regular galvanic cell problem however, then we multiply the standard reduction potential by -1. The formula E°cell=E°cathode−E°anode already accounts for this switch in the sign for E°anode. Hope this helped!

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

### Re: Test #2 Question 5B

You also don't have to reverse the equations because you're comparing all 3 of their reduction potentials. When reversing the equation and flipping the sign, you're not looking at its reduction potential anymore but its oxidation potential and you can't compare that to the reduction potentials of the other things you're looking at.