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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

Text Solution

Verified by Experts

Applying nernst equation to the given cell reaction,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Ag^(+)]^(2))=1.05V-(0.0591)/(2)"log"(0.160)/((0.002)^(2))=1.05-(0.0591)/(2)log(4xx10^(4))`
`=1.05-(0.0591)/(2)(4.6021)=1.05-0.14V=0.91V`
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Knowledge Check

  • Calculate emf of the cell in which the following reaction takes place :- Ni_(s) + 2Ag^(+) (0.002M) to Ni^(+2) (0.160M) + 2Ag_((s)) Given : E_("cell")^(o) = 1.05V, (2.303RT)/(F) = 0.06, log2 = 0.3

    A
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    B
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    C
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    D
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