## Test 2 #6 Ordering Reducing/Oxidizing Power

Christopher Tran 1J
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am

### Test 2 #6 Ordering Reducing/Oxidizing Power

Could someone let me know what they got for 6 a, b, and c for Test 3, and how they reasoned out the answers?

a. In order of increasing reducing power going from their neutral to second oxidation state: Pb, Cd, Mn
b. In order of increasing oxidation power going from their third oxidation state to neutral: U3+, Cr3+, Al3+
c. Please write the two species from parts a and b that would give the largest Ecell if assembled in an electrochemical cell.

Lauren Ho 2E
Posts: 70
Joined: Tue Oct 09, 2018 12:16 am

### Re: Test 2 #6 Ordering Reducing/Oxidizing Power

For part B. I got that the order of increasing oxidation power was Mg2+ < Ti2+ < Sn2+ because a more positive voltage means a stronger oxidizing agent. I don't know how to solve A and C, though :(

Chase Yonamine 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

### Re: Test 2 #6 Ordering Reducing/Oxidizing Power

To add, for part C you would just simply look at the biggest reducing and biggest oxidizing agent.

Chloe Likwong 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:23 am

### Re: Test 2 #6 Ordering Reducing/Oxidizing Power

Strongest reducing power: most negative value
Strongest oxidation power: most positive value

For C, find the most negative value (reducing power) as your anode because you'll be subtracting it in the formula [E°(cell) = E°(cathode) − E°(anode)] and the most positive value (oxidation power) as your cathode.

Hope this helps!