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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))` |
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Answer» 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) |
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