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Calculate the emf of the cell in which the following reaction takes place: Ni(s)+ 2Ag^+(0.002M) to Ni^2(0.160 M)+2Ag(s). Given that E_(cell)^0=1.05V |
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Answer» SOLUTION :`E_(cell)=E_(cell)^0-0.059/nlog[NI^(2+)]/[AG^+]^2` `=1.05-0.059/2log0.160/(0.002)^2=1.05-0.059/2log(4xx10^4)` `1.05-0.0295xx4.6021=1.05-0.135=0.914V` |
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