GS-2 Physical Sciences Question 49

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[sbains3528]

can someone explain this more clearly??

Which of the species in Table 1 is best suited to oxidize NO given that:

HNO2 + H+ + e- ↔ NO + H2O
Eo = +1.00 V

ANSWER: The more positive the Eo value, the greater the tendency for that half-reaction to proceed to the right. Since the Eo value for the Ce4+/Ce3+ equilibrium system is more positive than that for the HNO2/NO equilibrium system, the latter will be shifted to the left, that is, NO will be oxidized (= lose electrons; recall LEO is a GERC!; CHM 10.2).

[lrasja505901]

HNO2 gets reduced to NO E= +1.000
CE4+ gets reduced to CE3+ E= +1.695

CE4+ gets reduced. It is an oxidizing agent.
There it can oxidize NO. Only if it results in the reaction with an E greater than that of the initial rx.

1.6>1

[lrasja505901]

you were probably like me...

I thought Cd with its negative reduction potential was the best choice. That only means Cd tends to be oxidized. Not that it will oxidize something else..When looking to oxidize another species...we need an oxidizing agent, aka something that gets reduced.