Respuesta :
Answer: The electrode potential of the cell is 0.093 V
Explanation:
The given chemical cell follows:
[tex]Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)[/tex]
Oxidation half reaction: [tex]H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V[/tex]
Reduction half reaction: [tex]Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V[/tex]
Net cell reaction:
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}=0.13-(0.0)=0.13V[/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[Sn^{2+}]}{[Sn^{4+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.13 V
n = number of electrons exchanged = 2
[tex][H^{+}]=1.00M[/tex]
[tex][Sn^{2+}]=0.350M[/tex]
[tex][Sn^{4+}]=0.020M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})[/tex]
[tex]E_{cell}=0.093V[/tex]
Hence, the electrode potential of the cell is 0.093 V
