Answer:
The standard cell potential of the reaction is 1.24 Volts.
Explanation:
[tex]2Ag^+(aq)+Fe(s)\rightarrow 2Ag (s)+Fe^{2+}(aq)[/tex]
Reduction at cathode :
[tex]Ag^+(aq)+e^-\rightarrow Ag (s)[/tex]
Reduction potential of [tex]Ag^{+}[/tex] to Ag=[tex]E^o_{1}=0.80 V[/tex]
Oxidation at anode:
[tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]
Reduction potential of [tex]Fe^{2+}[/tex] to Fe=[tex]E^o_{2}=-0.44 V[/tex]
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.80 V -(-0.44 V)=1.24 V[/tex]
The standard cell potential of the reaction is 1.24 Volts.