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For the cell reaction 2Fe^(2+)(aq) + 2...

For the cell reaction
`2Fe^(2+)(aq) + 2I^(-)(aq) to 2Fe^(2+)(aq) + I_(2)`(aq)
`E_("cell"^(@) = 0.24 V` at 298 K. The standard Gibbs energy `(Delta_(r)G^(@))` of the cell reaction is:
[Given that Faraday constant F = 96500 C `"mol"^(-1)`]

A

`-46.32 kJ mol^(-1)`

B

`-23.16 kJ mol^(-1)`

C

`46.32 kJ mol^(-1)`

D

`23.16 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaG^(@) =-nFE_("cell")^(@)`
`=-2 xx 96500 xx 0.24`
`=-46320 J "mol"^(-1) = (-46320)/1000`
`=-46.32` kJ/mol
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For the cell reaction: 2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq) E_(cell)^(ɵ)=0.24V at 298K . The rstandard gibbs energy (triangle,G^(ɵ)) of the cell reaction is [Given that faraday constnat F=96400Cmol^(-1)]

(a) The cell in which the following reactions occurs: 2Fe^(3+)(aq)=2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F=96,500" C "mol^(-1) ) (b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F=96,500" C "mol^(-1) )

The cell in which the following reaction occurs 2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+2I_(2) " has "E_(cell)^(@)=0.236 V " at "298 K . Calculate standard Gibbs energy and equilibrium constant for the reaction.

The cell in which the following reaction occurs 2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(aq)+I_(2)(s) has E_(cell)^(0)=0.236V at 298 K. Calculate the standard gibbs energy and the equilibrium constant of the cell reaction.

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