Question

Calculate the voltage of the cell: zn 1 zn2+ (10m) 11 cu2+(0.20m) 1 cu

Answer 1

Answer: The voltage of the cell is 1.05 V.

Step-by-step explanation:

The given cell is:

`Zn(s)/Zn^(2+)(10M)||Cu^(2+)(0.20M)/Cu(s)`

Half reactions for the given cell follows:

Oxidation half reaction:
`Zn(s)\rightarrow Zn^(2+)(10M)+2e^-;E^o_(Zn^(2+)/Zn)=-0.76V`

Reduction half reaction:
`Cu^(2+)(0.02M)+2e^-\rightarrow Cu(s);E^o_(Cu^(2+)/Cu)=0.34V`

Net reaction:
`Zn(s)+Cu^(2+)(0.20M)\rightarrow Zn^(2+)(10M)+Cu(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)=0.34-(-0.76)=1.1V`

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+)])/([Cu^(2+)])`

where,

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

`E^o_(cell)` = standard electrode potential of the cell = +1.1 V

n = number of electrons exchanged = 2

`[Cu^(2+)]=0.02M`

`[Zn^(2+)]=10M`

Putting values in above equation, we get:

`E_(cell)=1.1-(0.059)/(2)* \log((10)/(0.2))\\\\E_(cell)=1.05V`

Hence, the voltage of the cell is 1.05 V.