Question

If the concentration of Sn2+ in the cathode compartment is 1.40 M and the cell generates an emf of 0.25 V , what is the concentration of Pb2+ in the anode compartment?

Answer 1

Answer : The concentration of
`Pb^(2+)` in the anode compartment will be,
`2.32* 10^(-9)M`

Solution :

The balanced cell reaction will be,

`Pb(s)+Sn^(2+)(aq)\rightarrow Pb^(2+)(aq)+Sn(s)`

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Tin (Sn) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

`E^0_([Sn^(2+)/Sn])=-0.14V`

`E^0_([Pb^(2+)/Pb])=-0.13V`

`E^0=E^0_([Sn^(2+)/Sn])-E^0_([Pb^(2+)/Pb])`

`E^0=-0.14V-(-0.13V)=-0.01V`

Now we have to calculate the concentration of
`Pb^(2+)`

Using Nernest equation :

`E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Pb^(2+)])/([Sn^(2+)]^2)`

where,

n = number of electrons in oxidation-reduction reaction = 2

`E_(cell)` = emf of the cell = 0.25 V

Now put all the given values in the above equation, we get concentration of
`Pb^(2+)`

`0.25=-0.01-(0.0592)/(2)\log ([Pb^(2+)])/(1.40)`

`[Pb^(2+)]=2.32* 10^(-9)M`

Therefore, the concentration of
`Pb^(2+)` in the anode compartment will be,
`2.32* 10^(-9)M`