Home
Class 12
CHEMISTRY
At equimolar concentration of Fe^(2+) an...

At equimolar concentration of `Fe^(2+) and Fe^(3+)` what must `[Ag^(+)]` be so that the voltage of the galvanic cell made from `Ag^(+)//Ag and Fe^(3+)//Fe^(2+)` electrodes equals zero? The reaction is `Fe^(2+) + Ag^(+) hArr Fe^(3+) + Ag`. Determine the equilibrium constant at `25^(@)C` for the reaction. Given `E_(Ag^(+)//Ag)^(@)= 0.799V and E_(Fe//Fe^(2+))^(@)= 0.771V`

Text Solution

Verified by Experts

The correct Answer is:
3
`E_("cell")^(@) = E_(Fe^(2+)//Fe^(3+))^(@) + E_(Ag^(+)//Ag)^(@)= -0.771 + 0.799= 0.028V`
At equilibrium, `E_("cell") =0`, so `0 = E_("cell")^(@) = (0.0591)/(1) "log" ([Fe^(3+)])/([Fe^(3+)][Ag^(+)])`
`E_("cell")^(@)= 0.0591 "log" (1)/([Ag^(+)]) rArr [Ag^(+)] = 0.34M` Now, log `K= (nE^(@))/(0.0591) rArr K= 3.0`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    BRILLIANT PUBLICATION|Exercise Level -III (Matching Column Type)|4 Videos

  • ELECTROCHEMISTRY

    BRILLIANT PUBLICATION|Exercise Level-III (Statement Type)|7 Videos

  • ELECTROCHEMISTRY

    BRILLIANT PUBLICATION|Exercise Level -III (Multiple Correct Answer Type)|9 Videos

  • D & F BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise Level -II|38 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL - II (ASSERTION - REASON) |5 Videos

Similar Questions

Explore conceptually related problems

Find the equilibrium constant at 25^@C for the reaction, Hg_2(NO_3)_2+2Fe(NO_3)_2=2Hg+2Fe(NO_3)_3 E_(Hg_2^(2+),Hg)^@=0.79 volt, E_(Fe^(3+),Fe^(2+))^@=0.77 volt

Electrometallurgy is used for the reduction of Cu Fe Na Ag

Assertion : In aqueous solution Cr^(+3) is more stable than Fe^(+3) Reason : E_(Cr^(+3))^(0)//Cr^(+2)=-0.41V , but E_(Fe^(3+)//Fe^(2+))^(0)=0.77V

The E at 25^(@) C for the following reaction is 0.22 V. The equilibrium constant at 25^(@) C is very H_(2)(g) +2AgCl(s) rarr -2Ag (s)+2HCl(aq)

Calculate the standard cell.potentials of galvanic cell in which the following reactions take place: Fe^2+(aq)+Ag+^*(aq)^* -> Fe^3+(aq)+Ag(s) Calculate the Delta r G^circ and equilibrium constant of the reactions.

A substance that will reduce Ag to Ag but will not reduce Ni^(2+) to Ni is : (E_(Ag^(+)//Ag)^(0))=0.80 V, E_(Ni^(2+)//Ni)^(0)

The ratio Fe^(3+) and Fe^(2+) ions in Fe_(0.9)S_(1.0) is.

E^@ for the reaction Fe+Zn^(2+)=Zn+Fe^(2+) is -0.35 V. The given cell reaction is

The equilibrium constant of the reaction 2Fe^(3+)+2I^(-)to2Fe^(2+)+I_(2) is 1xx10^(8) at 298 K. If E_(I_(2)//I^(-))^(@) is 0.54 V. what is E_(Fe^(3+)//Fe^(2+))^(@) ?