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Calculate the e.m.f. of the cell in which the following reaction takes place : `Ni(s) +2Ag^(+)(0.002 M)to Ni^(2+)(0.160 M)+2Ag(s)` Given `E_(cell)^(@)`=1.05 v |
Answer» `E_(cell)=E_(cell)^(@)-(0.0591)/(n)log[(Ni^(+2))]/([Ag^(+)]^(2)` `=1.05 -(0.0561)/(2)log(0.160)/(0.002^(2)` `=1.05-(0.0591)/(2)xx4.6` =1.05-0.14=0.91 V |
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