Question

The standard reduction potentials of cu2+ and ag+ in v are 0.337 and 0.799 . For what concentration of ag + will emf of the cell at 25 degree celsius will be zero if concentration of cu 2 + is 0.01 m

Answer 1

Answer : The concentration of $Ag^+$ will be, $1.47\times 10^{-9}M$

Solution :

The balanced cell reaction will be,  

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Cu^{2+}/Cu]}=0.337V$

$E^0_{[Ag^{+}/Ag]}=0.799V$

$E^0=E^0_{[Ag^{+}/Ag]}-E^0_{[Cu^{2+}/Cu]}$

$E^0=0.799V-(0.337V)=0.462V$

Now we have to calculate the concentration of $Ag^+$

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell  = 0

Now put all the given values in the above equation, we get  concentration of $Ag^+$

$0=0.462-\frac{0.0592}{2}\log \frac{(0.01)\times (1)^2}{(1)\times [Ag^+]}$

$[Ag^+]=1.47\times 10^{-9}M$

Therefore, the concentration of $Ag^+$ will be, $1.47\times 10^{-9}M$

Answer 2

Explanation:

Answer : The concentration of Ag^+Ag

+

will be, 1.47\times 10^{-9}M1.47×10

−9

M

Solution :

The balanced cell reaction will be,

Cu(s)+2Ag^{+}(aq)arrow Cu^{2+}(aq)+2Ag(s)Cu(s)+2Ag

+

(aq)arrowCu

2+

(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Cu^{2+}/Cu]}=0.337VE

[Cu

2+

/Cu]

0

=0.337V

E^0_{[Ag^{+}/Ag]}=0.799VE

[Ag

+

/Ag]

0

=0.799V

E^0=E^0_{[Ag^{+}/Ag]}-E^0_{[Cu^{2+}/Cu]}E

0

=E

[Ag

+

/Ag]

0

−E

[Cu

2+

/Cu]

0

E^0=0.799V-(0.337V)=0.462VE

0

=0.799V−(0.337V)=0.462V

Now we have to calculate the concentration of Ag^+Ag

+

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}E

cell

=E

cell

o

−

n

0.0592

log

[Cu][Ag

+

]

2

[Cu

2+

][Ag]

2

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell}E

cell

= emf of the cell = 0

Now put all the given values in the above equation, we get concentration of Ag^+Ag

+

0=0.462-\frac{0.0592}{2}\log \frac{(0.01)\times (1)^2}{(1)\times [Ag^+]}0=0.462−

2

0.0592

log

(1)×[Ag

+

]

(0.01)×(1)

2

[Ag^+]=1.47\times 10^{-9}M[Ag

+

]=1.47×10

−9

M

Therefore, the concentration of Ag^+Ag

+

will be, 1.47\times 10^{-9}M1.47×10

−9

M