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The e.m.f. of the cell obtained by combining `Zn` and `Cu` electrode of a Daniel cell with N calomel electrode in two different arrangements are `1.083V` and `0.018V` respectively at `25^(@)C`. If the standard reduction potential of N calomel electrode is `0.28V` find the emf og Daniel cell.

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Correct Answer - `E = 1.1V`
`E_(cell) = E_(ZN//Zn^(2+)) +E_(calomel(Red))`
`E_(ZN//Zn^(2+)) = 0.803`
`E_(cell) = E_(Cu^(2+)//Cu)`
`E_(cell) = 1.1V`

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