Question

What is the calculated value of the cell potential at 298K for an

Answer 1

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?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

Answer: The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.0-(-0.14)=0.14V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V