Which expression gives the value for `DeltaG^(@)` in kJ `mol^(-1)` for this reaction at `25^(@)C`?
`2Cr(s) +3Cu^(2+)(aq) to 2Cr^(3+) (aq) +3Cu(s) " " E^(@) = 0.43 V`

Options

A. `-6 xx 8.31 xx 0.43 xx 1000`
B. `(-6 xx 96500 xx 0.43 xx 1000)/(8.31)`
C. `(-6 xx 96500 xx 0.43)/( 1000)`
D. `(-6 xx 8.31 xx 0.43)/(1000)`

Answer

C

Explanation

`2Cr +3Cu^(2+) to 2Cr^(+3) +3Cu`
At Anode: `(Cr to Cr^(+3) +3e^(-)) xx 2`
At Cathode: `(Cu^(2+) +2e^(-) to Cu) xx 3`
No. of electron exchanged = 6
`DeltaG^(@) = -nFE_("cell")^(@)J//"mol"`
`DeltaG^(@) = -6 xx 96500 xx 0.43`
`DeltaG^(@)= (-6 xx 96500 xx 0.43)/(1000)`kJ/mol