Question

A voltaic cell is constructed that is based on the following reaction:

Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq).
If the concentration of Sn2+ in the cathode compartment is 1.00 M and the cell generates an emf of 0.16 V , what is the concentration of Pb2+ in the anode compartment?

Answer 1

Answer: The concentration of Pb2+ in the anode compartment is
`1.74* 10^(-6)M`

Step-by-step explanation:

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

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

`E^0=E^0_(cathode)- E^0_(anode)`

Where both
`E^0` are standard reduction potentials.

`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=(-0.14-(-0.13)V=-0.01V`

Now using Nernst Eqn :

`E=E^0-(0.059)/(n)\log([Pb^(2+)])/([Sn^(2+)])`

`0.16=(-0.01)-(0.059)/(2)\log([Pb^(2+)])/([1.00])`

`0.17=-0.0295\log([Pb^(2+)])/([1.00])`

`-5.76=\log([Pb^(2+)])/([1.00])`

`1.74* 10^(-6)=([Pb^(2+)])/([1.00])`

`[Pb^(2+)]=1.74* 10^(-6)`