A copper - silver cell is set up. The copper ion concentrations is `0.10` M. The concentration of silver ion is not known. The cell potential when measured was `0.422` V. Determine the concentration of silver ions in the cell.
`Given " "E^(@)Ag^(+)//Ag=+0.80 V,E^(@)Cu^(2+)//Cu=+0.34 V`

Explanation

(a). A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide `(PbO_(2))` as the cathode, and a `38%` solution of sulphuric acid `(H_(2)SO_(4))` as an electrolyte
When the battery is in use, the followig cell reaction take place
At anode: `Pb_((s))+SO_(4(aq))^(2-)toPbSO_(4(s))+2e^(-)`
At cathode: `PbO_(2(s))+SO_(4(aq))^(2-)+4H_((aq))^(-)+2e^(-)toPbSO_(4(s))+2H_(2)O_((l))`
the overall cell reaction is given by
`Pb_((s))+PbO_(2(s))+2H_(2)SO_(4(aq))to2PbSO_(4(s))+2H_(2)O_((l))`
When a battery is charged, the reverse of all these reactions take place.
Hence, on charging `PbSO_(4(s))` present at the anode and cathode is converted into `Pb_((s))` present at the anode and cathode is converted into `Pb_((s))` and `PbO_(2(s))` respectively (b). The given cell is
`underset(("anode"))(Cu||Cu^(2+))||underset(("cathode"))(Ag^(+)|Ag_((s))`)
the net reaction is
`Cu_((s))+2Ag_((ag))^(2+)to2Ag_((s))+Cu_((aq))^(2+)` ltbr. This involves transfer of 2 electrons Thus `n=2`
According to nearest equation`
`E_(cell)=E_(cell)^(0)+(0.059)/(2)log(([Cu][Ag^(+)]^(2))/([Cu^(2+)][Ag]^(2)))` ...(i)
`[Ag]=[Cu]=1` being solids
Also, `E_(Cell)=E_(Ag^(+)//Ag)^(0)-E_(Cu^(2+)//Cu)^(0)`
`=0.80-0.34`
`=0.46V`
Substituting the value of `E_(cell)^(0)` in equation (i), we get
`E_(cell)=0.46V+(0.059)/(2)log(([Ag^(+)]^(2))/([Cu^(2+)]))`
`0.422=0.46+(0.059)/(2)log(([Ag^(+)]^(2))/([0.10]))`
or `log(([Ag^(+)])/(0.10))=0.0515`
`[Ag^(+)]^(2)=0.005`
`Ag^(+)=0.07M`