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The E^(@) values for the changes given b...

The `E^(@)` values for the changes given below are measured against NHE at ` 27^(@) C `
`Cu^(2+) + e to Cu^(+) , E^(@) = + 0.15 V `,
`Cu^(+) + e to Cu, E^(@) = + 0.50 V `,
`Zn^(2+) + 2e to Zn , E^(@) = - 0.76 V `.
The temperature coefficient of emf a cell designed as `Zn|Zn^(2+) || Cu^(2+) || Cu` is ` - 1.4 xx 10^(-4)v` per degree . For a cell reaction in equilibrium `DeltaG= 0 and DeltaG^(@) = -2.303 RT log K_(c)`. The heat of reaction and entropy changes during the reaction are related by `DeltaG = DeltaH - TDeltaS`.
The decrease in free energy during the cell reaction in `Zn | Zn^(2+) (1M)||Cu^(2+)(1M)|Cu `, when its changes to ` 1M(Zn^(+2)) and 0.1M(Cu^(+2))` is .........

A

`2.037 xx 10^(5) J`

B

`2.116 xx 10^(5) J`

C

`2.037 xx 10^(6) J`

D

`2.116 xx 10^(6) J`

Text Solution

Verified by Experts

The correct Answer is:
A
`Cu^(+2) + e^(-) to Cu^(+) , 0.15 xx 1 , Cu^(+1) + 1e^(-) to Cu, 0.50 xx 1 `
` Cu^(+2) + 2e^(-) to Cu , E^(0) xx 2 , 2E^(0) =(0.50 + 0.15) implies E^(0) = 0.325 V `
For `Zn|Zn^(+2) ||Cu^(+2)|Cu`
`-0.76 " "+0.325`
`E_("cell")^(0) = 0.325 + 0.76 = 1.085V , DeltaG^(0) = nFE^(0) = -2 xx 96500 xx 1.085`
`DeltaG^(0) = -209 . 405 KJ = -2.095405 xx 10^(3)` joul
But , `DeltaG = ? " " DeltaG = DeltaG^(0) + RT log Q`
` -209. 405 + 2.303 xx (8.314 xx 298)/( 1000) log""((1)/(0.1)) = -209. 405 + 5.0706 = -203 . 700 KJ = -2.037 xx 10^(5) `joule
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