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For the cell Zn(s) | Zn^(2+)(2 M) ||Cu^(2+) (0.5 M) | Cu(s) (a) Write the equation for each half-reaction. (b) Calculate the cell potential at 25^(@) C. [Given, E_(Zn^(2+)//Zn)^(@) =-0.76 V, E_(Cu^(2+)//Cu)^(@) = +0.34 C] |
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Answer» Solution :`E^(@) =0.34 -(-0.76) = 1.1 V` `E_("cell") =E_("cell")^(@) -(0.0591)/2 LOG (ZN^(2+))/(Cu^(2+)) =1.1 -0.0591/2 log 2/0.5` `=1.1 -0.0295 log 4 = 1.1 -0.0295 xx 0.6021 = 1.1 - 0.0177 = 1.0823` V |
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