Question

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

electrochemical cell with the following reaction, when the H2
pressure is 6.56x10- atm, the H concentration is 1.39M, and
the Sn2+ concentration is 9.35x10-4M?
2H+ (aq) + Sn(s)—_H2(g) + Sn2+(aq)

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

Step-by-step 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)-(0.059)/(n)\log ([Sn^(2+)]* 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* 10^(-4)M`

`p_(H_2)=6.56* 10^(-2)atm`

Putting values in above equation, we get:

`E_(cell)=0.14-(0.059)/(2)* \log(((9.35* 10^(-4))* (6.56* 10^(-2)))/((1.39)^2))\\\\E_(cell)=0.273V`

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