In the following reactions first Identify the substance that is oxidized and the one that is reduced. Then identify the oxidizing and reducing agents. A. 2Li(s)+ F2(g) --> 2LiF(s) B. Fe(s) + CuSO4(aq) --> FeSO4(aq)+ Cu(s)
um there are different mnemonics to remember how to determine what is reduced and oxidized. my favorite is LEO and GER. Losing Electrons Oxidized and Gaining Electrons Reduction.
In a, bth the Li and the F start out with a neutral charge. When they react, the Li becomes positive and the F becomes negative. Becoming positive means losing electrons. Becoming negative means gaining electrons.
*both
So the Li is oxidized and the F is reduced.
Make sense?
not really
in this problem is there 2 parts to the anwser
Yes, I didn't get to the oxidation/reducing agents yet. Did you understand how I figured out which one is oxidized?
yes I do understand that part
Awesome! And did you understand which one is reduced?
yes because when you become negative you gain electrons=reduction
Perfect. The atom that causes the reduction is the reducing agent, so since the Li caused the F to be reduced it is the reducing agent.
And vice versa. The F is the oxidizing agent.
in b. would Fe be oxidized and CuSO be the reduction
Close. Fe is oxidized, but the element that is reduced is Cu.
The charge must change for it to be oxidized or reduced.
The SO4 is -2 the whole time.
so would the SO4 be the oxidizing agent?
No, Fe is the reducing agent and Cu is the oxidizing agent.
Whatever is oxidized is the reducing agent, and whatever is reduced is the oxidizing agent.
Oh well that does make sense but why is the SO4 not included?
Because it doesn 't gain or lose electrons.
Or cause anything else to gain or lose electrons.
that makes a lot of sense
I wish you were my chemistry teacher because I possibly would understand the material much better in lecture. Thank you
I'm glad I was able to help. :) I am a chem tutor and I do have the capability to do online lessons with a whiteboard. Message me if something like that would be helpful for you.
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