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Calculate the emf of the cell with the c...

Calculate the emf of the cell with the cell reaction
`Ni _((s))+2Ag ^(+)(0.002M)to Ni^(2+)(0.160M) +2 Ag_((s))`
`E_(cell)^(0)=1.05V.`

Text Solution

Verified by Experts

From the given cell reaction and Nernst equation
`E_(cell) =E_(cell)^(0) -(0.0591)/(n )log ""([Ni^(2+)])/([Ag^(+)]^(2))`
`=1.05 V- (0.0591)/(2)log ""([0.160])/([0.002]^(2))`
`=1.05 V-(0.0591)/(2)log (4xx 10^(4))`
`=1.05 -(0.0591)/(2)(4.6021)`
`=1.05 -0.14=0.91V=0.91V`
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