Question

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively. Calculate the potential for the following electrochemical cell: Zn(s)|Zn2+(0.125 M)||Ag+(0.240 M)|Ag(s)

Answer 1

Answer: The potential of the given cell is 1.551 V

Step-by-step explanation:

The given chemical cell follows:

`Zn(s)|Zn^(2+)(0.125M)||Ag^(+)(0.240M)|Ag(s)`

Oxidation half reaction:
`Zn(s)\rightarrow Zn^(2+)(0.125M)+2e^-;E^o_(Zn^(2+)/Zn)=-0.762V`

Reduction half reaction:
`Ag^(+)(0.240M)+e^-\rightarrow Ag(s);E^o_(Ag^(+)/Ag)=0.799V` ( × 2)

Net cell reaction:
`Zn(s)+2Ag^(+)(0.240M)\rightarrow Zn^(2+)(0.125M)+2Ag(s)`

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)=0.799-(-0.762)=1.561V`

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

`E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)])/([Ag^(+)]^2)`

where,

`E_(cell)` = electrode potential of the cell = ? V

`E^o_(cell)` = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

`[Zn^(2+)]=0.125M`

`[Ag^(+)]=0.240M`

Putting values in above equation, we get:

`E_(cell)=1.561-(0.059)/(2)* \log(((0.125))/((0.240)^2))`

`E_(cell)=1.551V`

Hence, the potential of the given cell is 1.551 V