Study the two half reactions of an electrochemical cell.
Ag(s) → Ag+(aq) + e–
E0 = 0.80 V
Cu(s) → Cu2+(aq) + 2e–
E0 = 0.34 V
Which option correctly gives the standard cell potential of
Ag+(aq) + Cu (s) → 2Ag(s) + Cu2+(aq) reaction?
-1.60
-0.46
1.60
0.46
Something is not right here.
First, the two equations you have written as oxidations have the wrong sign. Second, you don't give any options. Here is the way you get the cell potential for the one reaction you have written.
Ag+(aq) + Cu (s) → 2Ag(s) + Cu2+(aq)
2Ag^+(aq) + 2e ==> 2Ag(s) .......Eo red =+0.80
Cu(s) ==> Cu^2+(aq) + 2e................Eo ox = -0.34
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2Ag^+(aq) + Cu(s) ==> Cu^2+(aq) + 2Ag(s) Ecell = ?
Ecell = Eox + Ered = 0.80 + (-0.34) = ?
Repost if I've not anwered your question.
To determine the standard cell potential of the reaction:
Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)
you need to subtract the standard reduction potential of the Cu half-reaction from the standard oxidation potential of the Ag half-reaction.
The standard cell potential (E) is given by the formula:
E = E(cathode) - E(anode)
In this case, the Ag half-reaction is the cathode and the Cu half-reaction is the anode.
The standard reduction potential of the Cu half-reaction (E(cathode)) is 0.34 V.
The standard oxidation potential of the Ag half-reaction (E(anode)) is -0.80 V. (Remember, the given value is the reduction potential, so we need to take the negative of that value).
Now, we can substitute the values into the formula:
E = E(cathode) - E(anode)
E = 0.34 V - (-0.80 V)
E = 0.34 V + 0.80 V
E = 1.14 V
Therefore, the correct option is 1.14.
To find the standard cell potential of the reaction:
1. Write the two half reactions:
Ag(s) → Ag+(aq) + e– with E° = 0.80 V
Cu(s) → Cu2+(aq) + 2e– with E° = 0.34 V
2. Flip the second half reaction and change the sign of its potential:
Cu2+(aq) + 2e– → Cu(s) with E° = -0.34 V
3. Add the two half reactions to obtain the overall balanced equation:
Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)
4. Add the standard cell potentials of the two half reactions:
E°cell = E°(cathode) - E°(anode)
= 0.80 V - (-0.34 V)
= 0.80 V + 0.34 V
= 1.14 V
Therefore, the correct answer is 1.14 V.