Question

The `E^(@)` values corresponding to the following two reduction electrode processes are:
(i) `Cu^(+)//Cu=+0.52V`
(ii) `Cu^(2+)//Cu^(+)=+0.16V`
Formulate the galvanic cell for their combination. What will be the standard cell potential for it?
Calculate `Delta_(r)G^(@)` for the cell reaction `(F=96500" C "mol^(-1))`

Answer 1

For EMF to be +ve, oxidation should take place on electrode (ii) i.e., half-cell reactions will be
`Cu^(+)+e^(-)toCu`
`underline(" "Cu^(+)toCu^(2+)+e^(-)" ")`
Overall cell reaction: `2Cu^(+)toCu+Cu^(2+)`
Hence, the cell will be represented as: `Cu^(+)|Cu^(2+)||Cu^(+)|Cu`
`E_(cell)^(@)=E_(Red)^(@)(RHS)-E_(Red)^(@)(LHS)=0.52-0.16=0.36V`
`Delta_(r)G^(@)=-nFE_(cell)^(@)=-1xx96500" C "mol^(-1)xx0.36V=-34740" CV "mol^(-1)=-34740" J "mol^(-1)` (1 CV=1 J)