GS-1 Physical Sciences Question 49

[hellogoodb7913]

[jellywing_2058]

You are correct as the strongest reducing agent is the agent that is oxidized most.

If you look at the 4 different E° values for each of the reactions, the reduction of Cr3+ to Cr2+ gives the most negative value, which means that it would be the least reduced substance (least likely to gain an electron).

So, you need to reverse the reaction to its oxidized form in which Cr2+ loses an electron to form Cr3+ + e- with an inverse potential of 0.410.

Essentially what is important to note is that all the half reactions are in their reduction form and so the E° values are the potentials for the reductions of the half reactions. As such, the one with the lowest E° value is the half reaction that will be least likely to be reduced; it will therefore be the most likely to be oxidized.

So basically, the half reaction with the most negative (or smallest) E° potential value amongst all the electrochemical reactions will be the strongest reducing agent: answer choice B.

[y2boodz6039]

why is the answer Cr2+ instead of Cr3+?

"The reduced species of the electrochemical equilibrium with the most negative E'degree value is the strongest reducing agent"

what is the need of switching it to oxidation when in the table, they already show you the most negative E'degree value, leading to the conclusion that that half reaction will be the strongest reducing agent. (so you can easily cross out answers C and D)

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on the flip side, if it were to ask you the strongest oxidizing agent, would it be Mn2+ or MnO2?

according to the logic from the question, it would be Mn2+... but in this case, shouldn't it be MnO2?


thanks in advance.

[jellywing_2058]

It would be written as MnO2. The manganese as MnO2 has an oxidation state of +4 and it becomes reduced to +2 as Mn+2.

If you inverted the half reaction, it would be the oxidation of Mn+2 to Mn+4 and so the strongest oxidizing agent would be MnO2 - correct!