Answer:
The reduction potential of  [tex]Cu^2+/Cu[/tex] is  [tex]E_c^o = 0.34 \ V[/tex]
Explanation:
From the question we are told that
   The potential of the cell is  [tex]M_{cell} = E^o _{cell} = 1.100 \ V[/tex]
    The range is  [tex]R = 0.005 \ V[/tex]
    The temperature is  [tex]T = 25 ^oC[/tex]
Note the reason why Zn is oxized and  Cu is reduced is because Zn is higher than Cu on the electrochemical series
  The reaction at the anode is Â
       [tex]Zn ^{2-} _{(aq)} + 2e \to Zn_{(s)} \ \ \ \ \ E^o_a = -0.76 \ V[/tex]  Â
The [tex]E^o\ is \ the \ standard\ oxidation \ potential\ value\ for\ Zn\ oxidation[/tex]
  The reaction at the anode is Â
       [tex]Cu^{2+} _{(aq)} + 2e^{-} \to Cu_{(s)} \ \ \ \ \ \ \ \ E^o_c = c \ V[/tex] Â
Now
     [tex]E^o _{cell} = E_c^o - E_a^o[/tex] Â
substituting values
      [tex]1.100 = E_c^o -(- 0.76)[/tex]
      [tex]E_c^o = 1.100 - 0.76[/tex]
      [tex]E_c^o = 0.34 \ V[/tex]
Hence the reduction potential of  [tex]Cu^2+/Cu[/tex] is  [tex]E_c^o = 0.34 \ V[/tex]
