Study the two half reactions of an electrochemical cell.
Ag(s) → Ag^+(aq) + e^–
E^0 = 0.80 V
Cu(s) → Cu^2+(aq) + 2e^–
E^0 = 0.34 V
Which option correctly gives the standard cell potential of
Ag^+(aq) + Cu (s) → 2Ag(s) + Cu^2+(aq) reaction?
The answer I got is 0.46 is this correct?
no.Ag^+ + e ==> Ag is -0.80
Cu^2+ + 2e = Cu is +0.34
Add them. Your sign is wrong.
Oh, I see. Thank you, Dr. Bob.
To find the standard cell potential of the reaction, you can use the Nernst equation:
E(cell) = E(R) + E(L)
where E(cell) is the standard cell potential, E(R) is the reduction potential, and E(L) is the oxidation potential.
In this case, the reduction potential is for the reduction of Ag^+ ions to Ag:
Ag^+(aq) + e^– → Ag(s)
E(R) = +0.80 V
The oxidation potential is for the oxidation of Cu to Cu^2+ ions:
Cu(s) → Cu^2+(aq) + 2e^–
E(L) = -0.34 V
Now, we can substitute the values into the equation:
E(cell) = +0.80 V + (-0.34 V)
E(cell) = 0.46 V
So, the standard cell potential of the reaction Ag^+(aq) + Cu(s) → 2Ag(s) + Cu^2+(aq) is indeed 0.46 V. Your answer is correct.