Rank the following species in order of increasing reducing power going from their neutral to second oxidation state: Pb, Cd, Mn
What is the answer for this question? I got: Mn < Cd < Pb but I got it incorrect so what is the correct answer?
Test 2, Question 6 part a
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 89
- Joined: Fri Sep 28, 2018 12:16 am
- Been upvoted: 1 time
-
- Posts: 77
- Joined: Fri Sep 28, 2018 12:15 am
Re: Test 2, Question 6 part a
The species with the greatest reducing power has the most negative value, so the answer is:
Pb < Cd < Mn
Pb < Cd < Mn
-
- Posts: 27
- Joined: Mon Jun 25, 2018 3:00 am
Re: Test 2, Question 6 part a
Yeah... the whole reducing power/oxidizing power and reducing agent/oxidizing agent is a bit counterintuitive. Reducing power is how well the reaction oxidizes. We know that in an oxidation reaction we flip the reduction reaction and reduction potential, so our reduction potentials become the opposite sign, and hence become oxidation potentials. Therefore, the smaller the reduction potential, the greater the oxidation potential, and the greater the oxidation potential, the greater the reducing power!
-
- Posts: 89
- Joined: Fri Sep 28, 2018 12:16 am
- Been upvoted: 1 time
Re: Test 2, Question 6 part a
But because it says it goes from neutral to second oxidation, does that not mean you have to switch the signs?
-
- Posts: 64
- Joined: Wed Nov 15, 2017 3:04 am
Re: Test 2, Question 6 part a
You could also look at this by looking at electronegativity. The more electronegative it is the reducing agent and it itself is being oxidized.
Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”
Who is online
Users browsing this forum: No registered users and 3 guests