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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 vA. 1.05 VB. 0.912 VC. 1.19 VD. 2.05 V |
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Answer» Correct Answer - B `Ni_((S)) + 2Ag+ to Ni+2 + 2Ag_((S))` `Q=([Ni^(+2)])/([Ag^+]^2) = "0.16"/"(0.002)"^2="0.16"/(2xx10^(-3))^(2)="0.16"/(4xx10^(-6))` `Q=(0.16xx10^6)/4 = (16xx10^4) /4 =4xx104` `E_"cell" =E_"cell"^@ -0.059/n` log Q (From Nernst equation) `E_"cell"=1.05-0.06/2 log (4xx104)` = 1.05 – 0.03{log 4 + `log^(10)`4} = 1.05 – 0.03{0.6 + `4 log_(10)`} = 1.05 – 0.03{4.6} = 0.912 V |
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