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Calculate the equilibrium constant for t...

Calculate the equilibrium constant for the reaction
`Cu(s)+2Ag+(aq)rarrCu^(+2)(aq)+2Ag(s),E_("cell")^(@)=0.46V`.

A

`2.0 xx 10^(10)`

B

`4.0xx10^(10)`

C

`4.0xx10^(15)`

D

`2.4xx10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta G^(@)= n FE^(@)`
`= -2xx96500xx0.46= -88780 J`
Now `DeltaG^(@)= -2.303 "RT log"K_(c)`
`-88780 = -2.303xx8.314xx298 log K_(c)`
`log K_(c) = (88780)/(2.303xx8.314xx298)=15.56`
`K_(c)=3.63xx10^(15)` or `4.0xx10^(15)`
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