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The standard Gibbs free energy change (D...

The standard Gibbs free energy change `(DeltaG^(@)" in kJ mol"^(-1))`, in a Daniel cell `(E_("cell")^(@)=1.1 V)`, when 2 moles of `Zn (s)` is oxidized at 298 K, is closest to

A

`-212.3`

B

`-106.2`

C

`-424.6`

D

`-53.1`

Text Solution

Verified by Experts

The correct Answer is:
C
`Zn+Cu^(+2) rarr Zn^(+2)+Cu`
`2 Zn+2 Cu^(+2) rarr 2 Zn^(+2)+2 Cu`
For 2 moles of Zn, n=4
`DeltaG^(@)=-nFE_("Cell")^(@)=-4xx96500xx1.1=-424.6 kJ`
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