Based on the standard reduction potentials given below, which species is most likely to be oxidized?
Ag+ + e- → Ag, Eo= 0.80 V
Cu2+ + 2e- → Cu, Eo= 0.34 V
Zn2+ + 2e- → Zn, Eo= -0.76 V
Cd2+ + 2e- → Cd, Eo= -0.40 V
I thought the answer would be Zn2+ that'd be most likely to be oxidized because it has the lowest Standard reduction potential, but it was wrong and I'm not sure why.
Not Zn^2+. It's Zn and for the reason you cite. Zn ==> Zn^2+ + 2e has the highest oxidation potential
Well, it seems like you got caught in an electrifying situation! While it's true that the species with the lowest standard reduction potential is generally more likely to be oxidized, there is another factor to consider: the magnitude of the reduction potential.
In this case, the standard reduction potential of Zn2+ is indeed the lowest, but it is also quite far from the reduction potentials of the other species. The greater the difference between the reduction potentials of two species, the more likely the one with the higher potential will be reduced (opposite to oxidation).
So, to find the species that is most likely to be oxidized, we have to look for the one with the highest standard reduction potential. And in this list, the winner is Ag+ with a standard reduction potential of 0.80 V. Therefore, Ag+ is the one most likely to be oxidized. Keep up the good work, and don't let these electrifying questions shock you!
To determine which species is most likely to be oxidized, we need to consider the reduction potentials of the species involved. The species with the lowest reduction potential is most likely to be oxidized.
In this case, Zn2+ does have the lowest reduction potential of -0.76 V, but remember that reduction potentials are given as reduction reactions (gaining electrons) rather than oxidation reactions (losing electrons). To identify the species most likely to be oxidized, we need to consider the reverse reactions.
Let's reverse the reduction reactions:
Ag → Ag+ + e-, Eo = -0.80 V
Cu → Cu2+ + 2e-, Eo = -0.34 V
Zn → Zn2+ + 2e-, Eo = 0.76 V
Cd → Cd2+ + 2e-, Eo = 0.40 V
Now we can see that Ag has the lowest reverse reduction potential of -0.80 V, indicating that Ag+ is most likely to be reduced (meaning Ag is most likely to be oxidized). Therefore, Ag+ is the species most likely to be oxidized.
So, the correct answer is Ag+.
To determine which species is most likely to be oxidized, you need to consider the standard reduction potentials of the half-reactions involved.
A higher standard reduction potential indicates a greater likelihood for reduction to occur (gain of electrons), while a lower standard reduction potential indicates a greater likelihood for oxidation to occur (loss of electrons).
In this case, the species with the highest standard reduction potential is actually the most likely to be oxidized.
Let's compare the standard reduction potentials given for each species:
Ag+ + e- → Ag, Eo= 0.80 V
Cu2+ + 2e- → Cu, Eo= 0.34 V
Zn2+ + 2e- → Zn, Eo= -0.76 V
Cd2+ + 2e- → Cd, Eo= -0.40 V
Since Ag+ has the highest standard reduction potential of 0.80 V, it indicates that Ag+ is least likely to undergo oxidation (it is more likely to undergo reduction instead).
Zn2+ has the most negative standard reduction potential (-0.76 V), indicating that Zn2+ is the most likely to be oxidized (likely to lose electrons).
Therefore, the correct answer is Zn2+ is the species most likely to be oxidized.