Cu (s) , CuCl 2 (1.0 M) || Ag (s) , AgNO 3 (1.0 M) (Need help solving these ques

Question

Cu(s), CuCl2 (1.0 M) || Ag(s), AgNO3 (1.0 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.

Explanation / Answer

Cu(s), CuCl2 (1.0 M) || Ag(s), AgNO3 (1.0 M)

anode cathode

2) anode : At anode , oxidation takes place.

Cu ---------> Cu2+ + 2e-

cathode : At cathode, reduction takes place.

Ag+ + e- --------------> Ag

3) Overall balance reaction of electrochemical cell

Cu+ 2Ag+ ----------> Cu2+ + 2Ag

4)

  Cu ---------> Cu2+  + 2e- Eo = -0.33 V

Ag+ + e- --------------> Ag Eo = +0.8 V

Eocell = oxidation potential at anode + reduction potential at cathode

= -0.33 V + 0.8 V

= + 0.47 V

5)

Reaction quotient (Q) = [Cu2+] /[Ag+]2

6) Given that  [Cu2+] = 1.0 M

[Ag+] = 1.0 M

no of electrons transferred n = 2

Ecell = Eocell + 0.059/n log  [Cu2+] /[Ag+]2

= 0.47 - (0.059/2) log (1.0) / (1.0)2

= 0.47 -0

= + 0.47 V

Therefore,

Ecell = + 0.47 V

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