(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 t

Question

(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 t

(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)`)

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

(a) `2Fe^(3+)+2I^(-)to2Fe^(2+)+I_(2)(s)`
`E^(@)cell=0.236V,1F=96500C//mol`
`e^(-)+Fe^(3+)toFe^(2+)Jxx2`
`2I^(-)toI_(2)+2e^(-)`
`underlineoverline(" "2Fe^(3+)+2I^(-)+2e^(-)to2Fe^(2+)+I_(2)+2e^(-)" ")`
`thereforen=2`
`DeltaG^(@)=-nFE^(@)cell`
`=-2xx96500xx0.236`
`DeltaG^(@)=45548J`
(b) Quantity of electricity flowing through the wire
`=0.5xx2xx60xx60=3600C`
1F (96500C) is equivalent to flow of 1 mole o `e^(-)` i.e., `6.022xx10^(23)e^(-)s`
`therefore3600C` will be equivalent to `=(6.022xx10^(23))/(96500)xx3600=2.25xx10^(22)e^(-)`s