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The Gibbs energy change (in J) for the g...

The Gibbs energy change (in J) for the given reaction at `[Cu^(2+) ] = [Sn^(2+) ] = 1` M and 298 K is :
`Cu (s) + Sn^(2+) (aq) to Cu^(2+) (aq.) + Sn(s),`
`(E_(Sn^(2+) // Sn)^(0) = -0.16 V E_(Cu^(2+) |Cu)^(0) = 0.34 V ,` Take F = 96500 C `mol^(-1)` )

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Calculate the equilibrium constant for the reaction at 298K. Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.76V and E_(Cu^(2+)//Cu)^(@) = +0.34 V

Calculate the equilibrium constant for the reaction at 298K. Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.76V and E_(Cu^(2+)//Cu)^(@) = +0.34 V

E_(Cu^(2+)//Cu)^(@) =0.43V E_(Cu^(+)//Cu)^(@)=0.55V E_(Cu^(2+)//Cu^(+))^(@) =

The standard free energy change of the reaction Cu^(2+)+Sn(s)rarrCu(s)+Sn^(2+) Given : E^(@)=0.48V ) is:

The standard Gibbs energy for the given cell reaction in kJ mol^(-1) at 298 K is Zn(s) + Cu^(2+) (aq) to Zn^(2+)(aq) + Cu(s), E^@ = 2V " at " 298 K (Faraday's constant, F = 96000 C mol^(-1) )

E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

The standard Gibbs energy for the given cell reaction is KJmol^(-1) at 298 K is : Zn(s)+CU^(2+) (aq) rightarrow Zn^(2+) (aq) + Cu(s) E^(@)= 2V at 298 K (Friday's constant , F = 96000 C mol^-1 )

Calculate the standard free energy change in kJ for the reaction Cu^(+) +I^(-) to CuI Given: CuI + e to Cu+I^(-) E^(0)=-0.17V Cu^(+)+e to Cu E^(0)=0.53V

JEE MAINS PREVIOUS YEAR ENGLISH-JEE MAINS 2020-CHEMSITRY
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