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Explain the relationship between free energy of the cell and its emf. |
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Answer» Solution :The maximum work that can be OBTAINED from a galvanic CELL is `(W_(max))_(cell)=-nFE_(cell)"…(1)"` Here the (-) sign is introduced to indicate the work is done by the system on the surroundings. According to SECOND Law of thermodynamics, the maximum work done by the system is equal to the change in the Gibbs free ENERGY of the system. i,e, `W_(max)=DeltaG "...(2)"` From (1) and (2), `DeltaG=-nFE_(cell)"...(3)"` For a spontaneous cell reactions, the `DeltaG` should be negative. The above expression (3) indicates that `E_(cell)` should be positive to get a negative `DeltaG` value. When all the cell components are in their standard STATE, the equation becomes `DeltaG^(@)=-nFE_(Cell)^(@)` |
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