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For the cell Zn(s)|Zn^(2+) (2M)||Cu^(2+) (0.5 M)|Cu(s) (a) Write equation for each half reaction. (b) Calculate the cell potential at 25^@C |
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Answer» SOLUTION :(a) Half reaction OXIDATION half reaction: (i) `Zn(s) to Zn^(2+) (aq) +2e^(-)` Reduction half reaction: (II) `Cu^(2+) (aq) +2e^(-) to Cu(s) ` `E_(cell)^@=E_(cathode)-E_(anode)` `=+0.34-(-0.76)=1.1` `E_(cell)^@=E_(cell)^@-0.059/2log""(Products)/(Reactants)` `Zn_((s))+Cu^(2+) (0.5M)to Cu_((s))+Zn^(2+)(2M)` `E_(cell)^@=1.1-0.059/2 LOG""2/0.5` `=1.1-0.0295 times 0.6020` `=1.1 -0.0177` =1.082V |
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