Respuesta :
Answer: The cell potential is 0.712 V
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, silver will undergo reduction reaction will get reduced. Hydrogen will undergo oxidation reaction and will get oxidized
Half reactions for the given cell follows:
Oxidation half reaction: [tex]H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V[/tex]
Reduction half reaction: [tex]Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V[/tex] ( × 2 )
Net reaction: [tex]H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.80-(0.00)=0.80V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ag^{+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.80 V
n = number of electrons exchanged = 2
[tex][Ag^{+}]=0.0115M[/tex]
[tex][H^{+}]=0.355M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V[/tex]
Hence, the cell potential is 0.712 V
The cell potential of the cell is 0.71 V.
The overall equation of the redox reaction is;
2Ag^+(aq) + H2(g) ----> 2H^+(aq) + Ag(s)
The E°cell = E°cathode - E°anode
E°cell = 0.80 V - 0.00 V = 0.80 V
Using the Nernst equation;
Ecell = E°cell - 0.0592/n logQ
Where n = 2
Q = [H^+]^2/[Ag^+]^2 = (0.355 M)^2/(0.0115 M)^2
Q = 952.7
Substituting values;
Ecell = 0.80 V - 0.0592/2 log ( 952.7)
Ecell = 0.71 V
Learn more: https://brainly.com/question/5147266
