Answer
The correct is A)
Explanation:
A reduction half-reaction occurs in the cathode. The standard cell potencial (Eºcell) is higher than 0 for a spontaneous reaction. According to this, we deduce:
Eºcell = Eºr(cathode)-Eºr(anode) > 0 (spontaneous)
So, Eºr(cathode) > Eºr(anode) for a spontaneous reaction.
Since Eºr(cathode)= 0.80 V, we can say:
A) This half-reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential less than 0.80 V. TRUE
For example: Eºcell = Eºr(cathode)-Eºr(anode)= 0.80 V - (0.75 V)= 0.05 V > 0
B) This half-reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential greater than 0.80 V. FALSE
It is the opposite of A. For a value greater than 0.80 V, Eºcell will be <0 (non spontaneous)
C) This half reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential greater than -0.80 V. FALSE
This sentence is not true in all cases. For example: a reduction potential of 1.0 V is greater than -0.80 V, so:
Eºcell= Eºr(cathode)-Eºr(anode)= 0.80 V - (1.00 V)= -0.20 V < 0
D) It doesn't matter what half-reaction this is coupled with, it will proceed. FALSE, the reduction potential of the anodic reaction must be lower than 0.80 V.
