Question

10 points) Given the following two half-reactions, write the overall reaction in the direction in which it is product-favored, and calculate the standard cell potential.Pb2+ (aq) + 2 e–à Pb (s) Eº= –0.126 VFe3+ (aq) + e–à Fe2+ (s) Eº = +0.771 V

Answer 1

The over all reaction :

`2Fe^(3+)(aq)+Pb(s)\rightarrow 2Fe^(2+)(s)+Pb^(2+)(aq)`

The standard cell potential of the reaction is 0,.897 Volts.

Step-by-step explanation:

Reduction at cathode :

`Fe^(3+)(aq)+e^-\rightarrow Fe^(2+)(s)`..[1]

`E^o_(Fe^(3+)/Fe^(2+))=0.771 V`

Reduction potential of
`Fe^(3+)` to
`Fe^(2+)=0.771 V`

Oxidation at anode:

`Pb(s)\rightarrow Pb^(2+)(aq)+2e^-`.[2]

`E^o_(Pb^(2+)/Pb)=-0.126 V`

Reduction potential of
`Pb^(3+)` to
`Pb=-0.126 V`

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(red,cathode)-E^o_(red,anode)`

Putting values in above equation, we get:

`E^o_(cell)=E^o_(Fe^(3+)/Fe^(2+))-E^o_(Pb^(2+)/Pb)`

`=0.771 V-(-0.126 )=0.897 V`

The over all reaction : 2 × [1] + [2]

`2Fe^(3+)(aq)+Pb(s)\rightarrow 2Fe^(2+)(s)+Pb^(2+)(aq)`

The standard cell potential of the reaction is 0,.897 Volts.