Reduction Reaction E0 (V)
Cadmium (Cd) Cd2+ + 2e– → Cd -0.40
Silver (Ag) Ag+ + e– → Ag +0.80
If the standard cell potential of a cadmium-silver cell is 1.20 V and the reduction potential of Cd is –0.40 V, what is the oxidation potential of Ag?
-1.60
+1.60
-0.80
+0.80
I'm not sure why all of that information is given. I wonder if y ou've looked up the reduction of Ag. You can take the short cut. If the reduction potential of Ag^+ + e --> Ag is 0.80 then the oxidation potential is the reverse reaction written as an oxidation and change the sign to -0.80. However, if the reduction potential of the Ag^+/Ago is NOT given then the problem is done this way.
Cd ==> Cd^2+ + 2e .........Eox = + 0.40 volts
2Ag^+ + 2e --> 2Ag ..........Ered = ? volts
--------------------------------------------------
Cd + 2Ag^+ ==> 2Ag + Cd^2+ Ecell = 1.20 volts
So then you ask yourself if Eox + Ered = 1.20 v and Eox = 0.40, then
0.40 + Ered = 1.20 v and you solve for Ered.
r]pf[or
[agk3H[O
3
R;LREGK[p2KG;]RP
rl1\
[2gtjlfg
;WGL;q3k,WRLGH3'O3
To find the oxidation potential of Ag, we can use the equation:
Eºcell = Eºred(Cathode) - Eºred(Anode)
Given that the standard cell potential (Eºcell) is 1.20 V and the reduction potential of Cd (Eºred(Cd)) is -0.40 V, we can rearrange the equation to solve for the reduction potential of Ag (Eºred(Ag)):
Eºred(Ag) = Eºred(Cathode) - Eºcell
Eºred(Ag) = (+0.80 V) - (1.20 V)
Eºred(Ag) = -0.40 V
Therefore, the oxidation potential of Ag is -0.40 V.
To find the oxidation potential of Ag, we need to use the Nernst equation:
E(cell) = E(oxidation) + E(reduction)
We are given the following information:
E(cell) = 1.20 V
E(reduction) of Cd = -0.40 V
Let's substitute these values into the equation:
1.20 V = E(oxidation) + (-0.40 V)
Now we can solve for E(oxidation):
E(oxidation) = 1.20 V - (-0.40 V)
= 1.20 V + 0.40 V
= 1.60 V
So, the oxidation potential of Ag is +1.60 V.
Therefore, the correct answer is +1.60.