Fewpal
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in CBSE by (27 points)
Calculate ∆G° for the following cell:
Fe(s)│Fe2+(aq)(1M)║Cu2+(aq)(1M)│Cu(s). How is it related to the
equilibrium constant of the cell reaction Given E°Fe2+/Fe = -0.44 V , E°Cu2+/Cu = 0.34 V and 1F =
96500Cmol-1

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1 Answer

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ANSWER : - 150.54 KJ 

Given : 

Fe2+ | Fe = - 0.44 V

Cu2+ | Cu = 0.34 V 

Cell Reaction : 

Fe(s) | Fe2+(aq , 1M ) || Cu2+(aq , 1M )|| Cu(s)

To Find :

∆G° = ? 

Solution :

cell  = E°Cu2+ | Cu  - E°Fe2+|Fe

= 0.34 - ( - 0.44) 

= 0.34 + 0.44

= 0.78 V 

We know that , 

∆G° = - nFEcell

= - 2 × 96500 × 0.78 

= - 150540J 

= - 150.54 KJ 

\(\therefore\) ∆G° = - 150.54 KJ

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