Question

Pb2+(aq) + 2 e- --> Pb(s) Eo = -0.13V Zn2+(aq) + 2 e- --> Zn(s) Eo = -0.76V 9. Given the half-cell potentials above, when the reaction Zn(s) + Pb2+(aq) --> Zn2+(aq) + Pb(s) is made into a voltaic cell, the Ecell is: A. 0.63 V B. -0.63 V C. 0.89 D. 0.89

Answer 1

Answer: The
`E_(cell)` is 0.63 V

Step-by-step explanation:

In the given reaction :

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

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

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

`E^0_([Zn^(2+)/Zn])=-0.76V`

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

`E^0_(cell)=-0.13-(-0.76)=0.63V`

Thus the
`E_(cell)` is 0.63 V