If `E_(cell)^(@)` is 1.05 V, the emf of the cell for the following cell reaction
`Ni(s)+2Ag^(+)(0.004 M) rarr`
`Ni^(2+)(0.16M)+2Ag(s)` at 298 K in V is

Options

A. 0.932
B. 1.227
C. 0.732
D. 1.397

Answer

A

Explanation

Cell reaction
`underset((s))(Ni)+underset((0.004M))(2Ag^(+))rarr underset((0.16 M))(Ni^(2+))+underset((s))(2Ag)`
Given, `" "[Ag^(+]=0.004M`
`" "[Ni^(2+)]=0.16M`
`" "n=2`
`" "E_("cell")^(@)=1.05 V`
Apply in the Nernst equation,
`E_(cell)=E_(cell)^(@)-0.0591/n log""([ni^(2+)])/[Ag^(+)]^(2)`
Put the values in equation
`" "E_(cell)=1.05-0.0591/2log""([0.16])/[0.004]^(2)`
`" "E_(cell)=1.05-0.0591/2log""16/16 times 10^(4)`
`" "E_(cell)=1.05-0.059/2log10^(4)`
`" "E_(cell)=1.05-0.059/2""4 log10" "[because log10=1]`
`" "E_(cell)=1.05-0.059/2 times 4`
`" "E_(cell)=1.05-0.118 rArr E_(cell)=0.932V`