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प्रश्न
Calculate e.m.f. and ∆G for the following cell:
Mg (s) |Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu (s)
`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`
उत्तर
For the given cell representation, the cell reaction will be
Mg(s) + Cu2+(0.0001 M) → Mg2+(0.001 M) + Cu(s)
The standard emf of the cell will be given by
= 0.34 - (-2.37)
= 2.71 V
The Nernst equation for the cell reaction at 25 ºC will be
`E_(Cell)=E_(Cell)^@-(0.059/n)"log"([Mg^(2+)])/([Cu^(2+)])`
`=2.71-0.059/2"log"0.001/0.0001`
`=2.71-0.02955(log10)`
`=2.71-0.02955(1)`
`=2.68045V~~2.68V`
We know
∆G = −nFEcell
=−2×96500×2.68
=−517240 J mol−1
=−517.24 kJ mol−1
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