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

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.

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.