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1.	Calculate ∆G° for the following cell: Fe(s)│Fe2+(aq)(1M)║Cu2+(aq)(1M)│Cu(s). How is it related to theequilibrium constant of the cell reaction Given E°Fe2+/Fe = -0.44 V , E°Cu2+/Cu = 0.34 V and 1F = 96500Cmol-1
Answer» ANSWER : - 150.54 KJ Given : E°_{Fe²⁺\| Fe}= - 0.44 V E°_{Cu²⁺ \| Cu}= 0.34 V Cell Reaction : Fe_(s)\| Fe²⁺_{(aq , 1M )}\|\| Cu²⁺_{(aq , 1M )}\|\| Cu_(s) To Find : ∆G° = ? Solution : E°_cell = E°_{Cu²⁺ \| Cu} - E°_Fe²⁺\|Fe = 0.34 - ( - 0.44) = 0.34 + 0.44 = 0.78 V We know that , ∆G° = - nFE_cell = - 2 × 96500 × 0.78 = - 150540J = - 150.54 KJ \(\therefore\) ∆G° = - 150.54 KJ

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

Answer»

ANSWER : - 150.54 KJ

Given :

E°_{Fe²⁺| Fe}= - 0.44 V

E°_{Cu²⁺ | Cu}= 0.34 V

Cell Reaction :

Fe_(s)| Fe²⁺_{(aq , 1M )}|| Cu²⁺_{(aq , 1M )}|| Cu_(s)

To Find :

∆G° = ?

Solution :

E°_cell = E°_{Cu²⁺ | Cu} - E°_Fe²⁺|Fe

= 0.34 - ( - 0.44)

= 0.34 + 0.44

= 0.78 V

We know that ,

∆G° = - nFE_cell

= - 2 × 96500 × 0.78

= - 150540J

= - 150.54 KJ

\(\therefore\) ∆G° = - 150.54 KJ

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