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Calculate the equilibrium constant of th...

Calculate the equilibrium constant of the reaction `:`
`Cu(s)+2Ag(aq) hArrCu^(2+)(aq) +2Ag(s)`
`E^(c-)._(cell)=0.46V`

Text Solution

Verified by Experts

We know K=Antilog `[(nE^(@))/(0.059)]`
= Antilog `[(2xx0.46)/(0.059)]=3.92xx10^(15)`
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Calculate the equilibrium constant for the reaction Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s) Given that E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V

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Knowledge Check

  • Calculate the equilibrium constant of the reaction Cu(s)+2Ag^oplus rar Cu^+2(aq)+2Ag(s),E_cell^oplus=0.46V

    A
    `4xx10^15`
    B
    `3xx10^16`
    C
    `2xx10^13`
    D
    `156xx10^16`

  • The equilibrium constent of the reaction. Cu(s)+2Ag(aq). Leftrightarrow Cu^(2+) (aq.)+2Ag(s) E^(@)=0.46" V at 298 K is"

    A
    `2.0 xx 10^(10)`
    B
    `4.0 xx 10^(10)`
    C
    `4.0 xx 10^(15)`
    D
    `2.4 xx 10^(10)`

  • The equilbrium constant for the reaction : Cu + 2 Ag^(+) (aq) rarr Cu(2+) (aq) +2 Ag, E^@ = 0. 46 V at 299 K is

    A
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    B
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    C
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    D
    ` 1.212 V`

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