2. Write the half reactions that occur at the anode and cathode in this electroc
ID: 917014 • Letter: 2
Question
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
3. Write the overall balance reaction of the electrochemical cell.
4. Calculate Eocell of this electrochemistry cell. (include units)
5. Calculate the reaction quotient(Q) of this reaction
6. Calculate the expected Ecell for this reaction
I would really appreciate if someone could show me how to do 2-6 so I get the idea of how to do the rest of my problems because I have 10 very similar to this. I would greatly appreciate your help so much! Thank you in advance.
Explanation / Answer
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
Zn2+ + 2 e <--> Zn(s) 0.7618
Cu2+ + 2 e <--> Cu(s) +0.337
The most negative will be the one oxidizing (by rule)
Zn2+ + 2 e <--> Zn(s) 0.7618 --> oxidation, therefore cathode
Cu2+ + 2 e <--> Cu(s) +0.337 --> reduciton, therefore anode
3. Write the overall balance reaction of the electrochemical cell.
Invert the one oxidizing:
Cu2+ + 2 e <--> Cu(s) +0.337
Zn(s) <--> Zn2+ + 2 e +0.7618 (sign must be inverted)
Add both equations
Cu2+ + 2 e Zn(s) <--> Cu(s) + Zn2+ + 2 e
Cancel electrons
Cu2+ +Zn(s) <--> Cu(s) + Zn2+
4. Calculate Eocell of this electrochemistry cell. (include units)
The E°cell is given as:
E°cell = Ered + Eox
E°Cell = 0.337+0.7618 = 1.0988 V
5. Calculate the reaction quotient(Q) of this reaction
Q =[products]/[reactants]
Q = [Zn+2]/[Cu+2]
Q = 1/1 = 1
6. Calculate the expected Ecell for this reaction
Ecell = E°cell - 0.0592/n*logQ
n = number of electrons transferred, 2
Ecell = 1.0988 - 0.0592/2 * log(1)
Ecell = 1.0988 + 0
Ecell = 1.0988 V
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.