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Calculate the e.m.f. of the cell in whic...

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

A

1.05 V

B

0.912 V

C

1.19 V

D

2.05 V

Text Solution

Verified by Experts

The correct Answer is:
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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