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For the cell reaction: 2Fe^(3+)(aq)+2l...

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

`45.55 kJ//mol, 8.5xx10^(8)`

B

`50.12 kJ//mol, 7.2xx10^(7)`

C

`-50.12 kJ//mol, 8.5xx10^(8)`

D

`-45.55 kJ//mol, 9.6xx10^(7)`

Text Solution

Verified by Experts

The correct Answer is:
D
`2l^(-)rarr l_(2)+2e^(-)`
Hence, for the given cell reaction, n = 2.
`Delta_(r )G^(@)=-nF E_("cell")^(@)=-2xx96500xx0.236J`
`=-45.55 kJ mol^(-1)`
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(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)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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