Respuesta :
The question is incomplete, here is the complete question:
What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?
[tex]2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)[/tex]
Answer: The cell potential of the given electrochemical cell is 0.273 V
Explanation:
For the given chemical equation:
[tex]2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)[/tex]
The half cell reactions for the given equation follows:
Oxidation half reaction: [tex]Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V[/tex]
Reduction half reaction: [tex]H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V[/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}=0.0-(-0.14)=0.14V[/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{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.14 V
n = number of electrons exchanged = 2
[tex][H^{+}]=1.39M[/tex]
[tex][Sn^{2+}]=9.35\times 10^{-4}M[/tex]
[tex]p_{H_2}=6.56\times 10^{-2}atm[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V[/tex]
Hence, the cell potential of the given electrochemical cell is 0.273 V
