Question

The reduction potentials of `Cu^(2+)//Cu` and `Ag^(+)//Ag` electrodes are 0.34V and 0.80 v respectively. For what concentration of `Ag^(+)` ions will the EMF of the cell at `25^(@)C` is zero ? Given that the concentration of `Cu^(2+)` is 0.01 M.

Answer 1

The redox reaction taking place in the cell is
`Cu(s)+2Ag^(+)(aq) to Cu^(2+)(aq)+2Ag(s)`
According to Nernst equation :
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cu^(2+)(aq)])/([Ag^(+)(aq)]^(2))`
`E_(cell)=0, E_(cell)^(@)=E_((Ag^(+)//Ag))^(@)-E_((Cu^(2+)//Cu))^(@)=0.80-0.34=0.46" V"`,
`[Cu^(2+)(aq)]=0.01 M`
Substituting the values in the equation,
`0=0.46-(0.0591)/(2)"log"((0.01))/([Ag^(+)(aq)]^(2))`
`"log"(0.01)/([Ag^(+)(aq)]^(2))=(2xx0.46)/(0.0591)=15.567`
`(0.01)/([Ag^(+)(aq)]^(2))="antilog "15.567=3.69xx10^(15)`
`[Ag^(+)(aq)]^(2)=(0.01)/(3.69xx10^(15))=2.71xx10^(-17)`
`[Ag^(+)(aq)]=(2.71xx10^(-17))^(1//2)=(27.1xx10^(-18))^(1//2)=5.21xx10^(-9)M`.