____________ A voltaic cell has an E_cell^degree of 0.23 V under standard condit
ID: 527175 • Letter: #
Question
____________ A voltaic cell has an E_cell^degree of 0.23 V under standard conditions. What can you conclude K_eq for the reaction? Keq 1 K_eq = 0 K_eq = 1 K_eq = E_cell^degree ___________ You want to make a voltaic cell with copper (Cu, Cu^2+) and another material. Which material do you choose to maximize E_cell^degree? Is copper the anode or cathode? Ag^+/Ag, copper is anode Ag^+lAg, copper is cathode Zn^2+/Zn, copper is anode Zn^2+/Zn, copper is cathode Not enough information ________ Which conditions provide the highest cell potential (E_cell) in a concentration cell? [anode] = [cathode] uparrow [anode], downarrow[cathode] downarrow[anode], downarrow[cathode] uparrow[anode], downarrow[cathode] downarrow[anode], uparrow[cathode]Explanation / Answer
4)
since Eocell is positive, the reaction is spontaneous and hence equilibrium constant will be greater than 1
Answer: B) Keq > 1
5)
when we used Zn:
from data table:
Eo(Zn2+/Zn(s)) = -0.762 V
Eo(Cu2+/Cu(s)) = 0.337 V
the electrode with the greater Eo value will be reduced and it will be cathode
here:
cathode is (Cu2+/Cu(s))
anode is (Zn2+/Zn(s))
The chemical reaction taking place is
Cu2+ + Zn(s) --> Cu(s) + Zn2+
Eocell = Eocathode - Eoanode
= (0.337) - (-0.762)
= 1.099 V
when we use Ag:
from data table:
Eo(Cu2+/Cu(s)) = 0.337 V
Eo(Ag+/Ag(s)) = 0.8 V
the electrode with the greater Eo value will be reduced and it will be cathode
here:
cathode is (Ag+/Ag(s))
anode is (Cu2+/Cu(s))
The chemical reaction taking place is
Ag+ + Cu(s) --> Ag(s) + Cu2+
Eocell = Eocathode - Eoanode
= (0.8) - (0.337)
= 0.463 V
Clearly Eo is larger in case of Zn. There copper was used as cathode
Answer: D
6)
E = Eo - (2.303*RT/nF) log Q
Q = [anode] / [cathode]
Greater [Cathode] implies smaller Q and hence larger E
Answer: E
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