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…

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asked Feb 16, 2021 61.6k views

1 Answer

Vsasv · Answer 1 · 2021-02-20T16:44:44+0000

Answer:

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.