Question
Please help. I need to calculate the emf for each of thefollowing cells, writing the overall reaction, identifying theanoda/cathode and stating the metals (zinc, copper, lead) in orderof decreasing ease of oxidation (most easily oxidized first). 1. Zn/Zn2+ and Pb/Pb 2+ 2. Zn/Zn2+ and Cu/Cu2+ 3.Cu/Cu2+ and Pb/Pb2+ Please help. I need to calculate the emf for each of thefollowing cells, writing the overall reaction, identifying theanoda/cathode and stating the metals (zinc, copper, lead) in orderof decreasing ease of oxidation (most easily oxidized first). 1. Zn/Zn2+ and Pb/Pb 2+ 2. Zn/Zn2+ and Cu/Cu2+ 3.Cu/Cu2+ and Pb/Pb2+
Explanation / Answer
1) Zn (s) + Pb+2 (aq) ------> Pb(s) + Zn+2 (aq) Therefore Oxdation half cell : Zn ------>Zn+2 +2e Reduction half cell : Pb+2 + 2e --------> Pb(s) Thus Eocell =Eocathode - Eoanode = - 0.13V - (-0.76V) =0.63 V 2) Zn (s) +Cu+2 (aq) ------> Cu (s) +Zn+2 (aq) Therefore Oxdation half cell : Zn ------>Zn+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.76V) =1.10 V 3) Pb (s) +Cu+2 (aq) ------> Cu (s) +Pb+2 (aq) Therefore Oxdation half cell : Pb ------>Pb+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.13 V) = 0.47V = - 0.13V - (-0.76V) =0.63 V 2) Zn (s) +Cu+2 (aq) ------> Cu (s) +Zn+2 (aq) Therefore Oxdation half cell : Zn ------>Zn+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.76V) =1.10 V 3) Pb (s) +Cu+2 (aq) ------> Cu (s) +Pb+2 (aq) Therefore Oxdation half cell : Pb ------>Pb+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.13 V) = 0.47V Therefore Oxdation half cell : Zn ------>Zn+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.76V) =1.10 V 3) Pb (s) +Cu+2 (aq) ------> Cu (s) +Pb+2 (aq) Therefore Oxdation half cell : Pb ------>Pb+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.13 V) = 0.47V = 0.34V - (-0.76V) =1.10 V 3) Pb (s) +Cu+2 (aq) ------> Cu (s) +Pb+2 (aq) Therefore Oxdation half cell : Pb ------>Pb+2 +2e Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.13 V) = 0.47V Reduction half cell : Cu+2 + 2e --------> Cu(s) Thus Eocell =Eocathode - Eoanode = 0.34V - (-0.13 V) = 0.47V = 0.34V - (-0.13 V) = 0.47V