 
    e) See your teacher for solution
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Solution
Cell emf = highest (half cell emf)- lowest (half cell emf)
a) Copper half cell = 0.34V
       Hydrogen half cell = 0.00V
      Cell emf = 0.34 -0.00V = 0.34V
b) Copper half cell = 0.34V
       Iron half cell = -0.41V
      Cell emf = 0.34 - - 0.41V = 0.34 + 0.41 = 0.75V
c) 
      Iron half cell = -0.41V
       Hydrogen half cell = 0.00V
      Cell emf = 0.00V --0.41= 0.00 + 0.41V = 0.41V
d) 
      Permanganate ion = 1.49
       Hydrogen half cell = 0.00V
      Cell emf = 1.49 -0.00V = 1.49V
e) 
      Magnesium half cell = -2.38V
       Iron half cell = -0.41V
      Cell emf = -0.41 - -2.38V = -0.41 + 2.38 = 1.97V
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Remember. Oxidation always occurs at the anode (-) while reduction occurs at the cathode(+).
a) 
     and
 and 
    Reduction = cathode (+) ------------------------ 
 
    
Oxidation 
    = anode (-) ---------------------------- 
b)  and
 
    and 
Reduction 
    = cathode (+) ------------------------ 
 
    
Oxidation 
    = anode (-) ---------------------------- 
c) and
 
    and 
Reduction = cathode (+) ------------------------ 2H+(aq) + 2e <=> H2(g)
Oxidation 
    = anode (-) ---------------------------- 
d) and
and 
    
Reduction 
    = cathode (+) -----------------
Oxidation 
    = anode (-) ---------------------------- 
e) and
and 
    
Reduction 
    = cathode (+) ------------------------
Oxidation 
    = anode (-) ---------------------------- 
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| Potential difference. Exercises | 
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| 1) Using the table above predict if a reaction will occur and write the reaction when: a) hydrogen gas is bubbled 
        through a solution of copper nitrate. Solution 2) Using the table above calculate the emf of the cell created when the following half cells are connected. a)  b)  c)  d)  e)  Solution 3) In question 2) above indicate 
          the reaction occuring at the anode and at the cathode. Indicate the 
          direction of electron flow. Solution |