Question

Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the following reaction, where zinc is the anode and lead is the cathode.

Zn(s) + Pb2+(aq) -> Zn2+(aq) + Pb(s)

Half-reaction: Standard Reduction Potential:

Zn2+(aq) + 2e- -> Zn(s)= -0.763

Pb2+(aq) + 2e- -> Pb(s)= -0.126

a. -0.889 V

b. +0.889 V

c. +0.637 V

d. +1.274 V

e. -0.637 V

Answer 1

Answer: C) 0.637 V

Step-by-step explanation:

The balanced redox reaction is:

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

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead 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.

`Zn^(2+)(aq)+2e^-\rightarrow Zn(s)`= -0.763

`Pb^(2+)(aq)+2e^-\rightarrow Pb(s)`= -0.126

`E^0_([Zn^(2+)/Zn])=-0.763V`

`E^0_([Pb^(2+)/Pb])=-0.126V`

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

`E^0=-0.126-(-0.763V)=0.637V`

The standard emf of a cell is 0.637 V