Question

Zn(s), Zn(NO3)2 (0.3 M) || Cu(s), CuCl2 (0.5 M)

(Need help solving these questions, so I can solve the rest of the questions like this in my lab. It is greatly appreciated. Thank you)

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 this electrochemical cell.
4. Calculate E°cell of this electrochemical cell. (include units)
5. Calculate the reaction quotient (Q) of this reaction.
6. Calculate the expected Ecell for this reaction

Answer 1

See Explanation

Step-by-step explanation:

At the anode;

Zn(s) -----> Zn^2+(aq) + 2e

At the cathode;

Cu^2+(aq) + 2e ------> Cu(s)

Overall electrochemical reaction;

Zn(s) + Cu^2+(aq) ------> Zn^2+(aq) + Cu(s)

E°cell = E°cathode - E°anode

E°cell = 0.34 - (-0.76)

E°cell = 1.1 V

Q = [0.3 M]/[0.5 M]

Q = 0.6

From Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Ecell = 1.1 - 0.0592/2 log (0.6)

Ecell = 1.1 - 0.0296 log (0.6)

Ecell = 1.11 V