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Solutions

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) | Al+3  (aq, 1M) || Cu+2 (aq, 1 M) | Cu (s)
 
 

2. Write the cell diagram and calculate the Eo for the Zn/ Zn+2 and Ca/Ca+2 half-cells.

 Eo= 2.11 V
cell diagram:
Ca (s) | Ca+2 (aq, 1M) ||  Zn+2 (aq, 1 M) Zn (s)

 

 

 

3. a) Will I- react with Br2 Yes
  b) Will Ni+2 react with Br2? No
  c) Will Ni react with Br2? Yes
  Will Co+2 react with Br2 to give Co+3? No, reacts with Co+2 to give Co (s)

 

4. Rank the following ions from strongest to weakest oxidizing agents:

   
Eo , V
Ag+
strongest
+ 0.80
Pb+2
 
- 0.13
Al+3
 
- 1.66
Sr+2
 
- 2.89
Li+
weakest
- 3.05

 

5. 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

 

 

6. Is the reaction below spontaneous?  no Calculate the EoEo = - 0.47 V

Pb+2  +  Cu  ------>    Pb    +   Cu+2

 

 

7. Determine the \(\Delta{G}^o\) for the Ag+/Ag and Sn+2/Sn half-cells.

missing...anyone want to do this?

 

8. Calculate \(\Delta{G}^o\) for Cu+2/Cu and Mn+2/Mn half-cells.  Is this reaction spontaneous?

Eo = 1.52 V;

\(\Delta{G}^o = -293,360 J\)

or

\[- 293.4 \, kJ\]

Yes, this reaction is spontaneous as written.


5. Calculate \(\Delta{G}\) for Ni+2/Ni and Fe+2/Fe half-cells if the [Ni+2] = 0.25 M and the [Fe+2] = 0.50 M.

Eo = 0.19 V; E = 0.181 V; DG = - 34956 J or - 35.0 kJ

6. Determine the \(\Delta{G}\) for the Al+3/Al and Ni+2/Ni half-cells.if the [Al+3] = 0.50 M and the [Ni+2] = 0.25 M.

Eo = 1.41 V; E = 1.398 V; \(\Delta{G}\) = - 809442 J or - 809.4 kJ