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The emf of the cell reaction, Zn((s))+...

The emf of the cell reaction,
`Zn_((s))+Cu_((aq))^(+2) to Zn_((aq))^(+2) +Cu_((s))`
Calculate the entropy change. Given that enthalpy of the reaction is -216.7 KJ mol`""^(-1) And E_(Zn^(+2)//Zn)^(0)=-0.76 V and E_(Cu^(+2)//Cu)^(0)=+0.34V`

Text Solution

Verified by Experts

The correct Answer is:
`-14.76 JK^(-1)mol^(-1)`
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Calculate the equilibrium constant K for the reaction at 298 K Zn(s) + Cu^(2+)(aq) [Given: E_(Zn^(2+)//Zn)^(@) =-0.76 V, E_(Cu^(2+)//Cu)^(@) = +0.34 V ]

The emf of the cell reaction Zn(s) +Cu^(2+) (aq) rarr Zn^(2+) (aQ) +Cu(s) is 1.1V . Calculate the free enegry change for the reaction. If the enthalpy of the reaction is -216.7 kJ mol^(-1) , calculate the entropy change for the reaction.

Knowledge Check

  • For the cell reaction Zn_((s)) + Cu_("0.1 M")^(2+) to Zn_(0.1 M)^(2+) + Cu_((s)) if the standard EMF of the cell is E^(@) , then

    A
    `E gt E^(@)`
    B
    `E lt E^(@)`
    C
    `E = E^(@)`
    D
    `E le E^(@)`

  • Calculate the maximum work that can be obtained from the decimolar Daniell cell at 25^(@)C . Given E^(c-)._((Zn^(2+)|Zn))=-0.76V and E^(c-)._((Cu^(2+)|Cu))=0.34V

    A
    `193.0k J`
    B
    `212.3kJ`
    C
    `81.06kJ`
    D
    `40.53kJ`

  • For the cell reaction, Cu_(C_(2))^(2+)(aq)+Zn(s) to Zn_(C_(1))^(2+)(aq)+Cu(s) The change in free energy (DeltaG) at a given temperature is a function of :

    A
    In `C_(1)`
    B
    In `(c_(2)//c_(1))`
    C
    In `(c_(1)+c_(2))`
    D
    In `c_(2)`

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