Arkansas State University
Department of Chemistry
and Physics
Answers
Electrochemistry

1. Write the cell diagram for the Cu+2/Cu and Al+3/Al half-cells and calculate the Eocell for the Cu+2/Cu and Al+3/Al half-cells.
 
Write half-reactions with more negative on top:
Al+3  (aq) + 3 e-  ------->   Al (s)
Eo = -1.66 V
Cu+2 (aq) + 2 e-  ------->  Cu (s)
Eo = +0.34 V
Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of Eo
Al (s)   ------->    Al+3  (aq) + 3 e-
Eo = +1.66 V       (anode)
Cu+2 (aq) + 2 e-  ------->  Cu (s)
Eo = +0.34 V       (cathode)
   
Add cell potentials of half-reactions to determine Eocell Eocell = +2.00 V

cell diagram (anode first):

Al (s)I Al+3  (aq, 1M)I KCl (sat'd) I Cu+2 (aq, 1 M) I Cu (s)

2. Determine the E for the Ag+/Ag and Sn+2/Sn half-cells if the [Ag+] = 1.0 M and the [Sn+2] = 0.25 M.
 
Write half-reactions with more negative on top:
Sn+2  (aq) + 2 e-   ------->   Sn (s)
Eo = -0.14 V
Ag+ (aq) + 1 e-  ------->  Ag (s)
Eo = +0.80 V 
Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of Eo
Sn (s)   ------->    Sn+2  (aq) + 2 e-
Eo = +0.14 V       (anode)
Ag+ (aq) + 1 e-  ------->  Ag (s)
Eo = +0.80 V      (cathode)
   
Add cell potentials of half-reactions to determine Eocell Eocell = +0.94 V
   
To determine E, use Nernst equation: E = Eo-(0.0591/n)log Q
E = Eo-(0.0591/n)log Q; n = 2 (make e- lost = e- gained in the half-reactions)
Q =  [Sn+2]/ [Ag+]2([Sn+2] is on the product side, [Ag+] is on the reactant side after switching half-reactions)
E = 0.94V -(0.0591 V/2)log {0.25M Sn+2/ (1.00 Ag+)2} E = 0.958 V

3. Determine the DGo for the Ag+/Ag and Sn+2/Sn half-cells.

DGo = - nFEo   = - 2 mol( 96500 J/V mol)(0.94 V) = -181,420 J = -181.42 kJ