Home
Class 12
CHEMISTRY
Calculate the equilibrium constant for t...

Calculate the equilibrium constant for the reaction.
`Fe(s) + Cd^(2+) (aq) hArr Fe^(2+)(aq) + Cd(s)`
[ Given, `E_(Cd^(2+) //Cd)^(@) = - 0.40V , E_(Fe^(2+)//Fe)^(@) = - 0.44 V `]

Text Solution

Verified by Experts

`E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@)`
`= - 0.40 - ( -0.44)`
`= -0.40 + 0.44 = 0.04V`
`E_("cell")^(@) = (0.059)/( n ) log K_(C )`
`E_("cell")^(@) = ( 0.059)/( 2) log K_(C ) = 0.04 `
or log `K_(C ) = ( 2 xx 0.4)/( 0.059) = 1.356 `
`K_(C ) = ` antilog 1.356 = 22.70
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE (PART 1) ( CONDUCTOR ELECTROLYTES AND THEIR DISSOCIATION, ELECTROLYSIS, ITS LAW AND APPLICATIONS)(MULTIPLE CHOICE TYPE QUESTIONS) |4 Videos

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE (PART 1) ( CONDUCTOR ELECTROLYTES AND THEIR DISSOCIATION, ELECTROLYSIS, ITS LAW AND APPLICATIONS)(VERY SHORT ANSWER TYPE QUESTIONS) |8 Videos

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise SAMPLE QUESTION (PART II)|3 Videos

  • D-BLOCK ELEMENTS

    ARIHANT PUBLICATION|Exercise Chapter practice (Long answer type questions)|2 Videos

  • ELEMENTS : NITROGEN FAMILY

    ARIHANT PUBLICATION|Exercise CHAPTER PRACTICE ( Long Answer Type Questions ) |11 Videos

Similar Questions

Explore conceptually related problems

Find the equilibrium constant for the reaction, Cu^(2+) + ln^(@) hArr Cu^(+) + ln ^(3+) Given that, E_(Cu^(2+)//Cu^(+)) = 0.15V , E_(ln^(2+) //ln^(+))^(@) = -0.4 V, E_(ln^(3+) //ln)^(@) = 0.42V

The equilibrium constant for the reaction Cu_(s) + 2Ag^+ _ (aq) rarr Cu^2+ _ (aq) +2Ag at 298 k is Given E^@cell=0.46v

Calculate the standard cell potential of a galvanic cell in which the following reaction takes place 2Cr(s) + 3Cd^(2+) ( aq) rarr 2Cr^(3+) ( aq) + 3Cd(s) Calculate Delta_( r) G^(@) and equilibrium constant K of the above reaction at 25^(@)C . [Given, E_(Cr^(3+) //Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.40V, 1F = 96500 C "mol"^(-1)]

Calculate the emf for the given cell at 25^(@)C . Cr| Cr^(3+) ( 0.1M) || Fe^(2+) ( 0.01M) | F e [Given, E_(Cr^(3+)//Cr)^(@)= -0.74V, E_(Fe^(2+)//Fe)^(@) = -0.44V ]

Calculate the value of equilibrium constant for the reaction at 298 k. Cu(s) +2Ag^+(aq)rarrCu^(2+)(aq)+2Ag(s) E^@ (Ag2+/Ag) =0.80V and E^@( Cu2+/Cu ) =0.34V

Calculate the standard emf of the cell having the cell reaction. Zn(s) + Co^(2+) ( aq) rarr Zn^(2+) ( aq) + Co(s) E_(Zn//Zn^(2+)) ^(@) = 0.76V, E_(Co//Co^(2+))^(@) = 0.25V

Calculate the cell emf and Delta_(r ) G^(@) for the cell reation at 25^(@)C . Zn(s) | Zn^(@+) ( 0.1 M )|| Cd^(2+) ( 0.01M) | Cd(s) Given, E_(Zn^(2+) //Zn)^(@) = - 0.763V , E_(Cd^(2+) //Cd) ^(@) = - 0.403 V 1F = 96500 C "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) ], Find E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) then Delta _(r ) G^(@) by using formula, Delta _(r ) G^(@) = - n FE_("cell")^(@)