The values of `E^(@)` of some of the reactions are given below : `{:(I_(2)+2e^(-) rarr 2I^(-),,E^(@)=+0.54" volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=

Question

The values of `E^(@)` of some of the reactions are given below : `{:(I_(2)+2e^(-) rarr 2I^(-),,E^(@)=+0.54" volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=

The values of `E^(@)` of some of the reactions are given below :
`{:(I_(2)+2e^(-) rarr 2I^(-),,E^(@)=+0.54" volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=+1.36" volt"),(Fe^(3+)+e^(-) rarr Fe^(2+),,E^(@)=+0.76" volt"),(Ce^(4+)+e^(-) rarr Ce^(3+),,E^(@)=+1.60" volt"),(Sn^(4+)+2e^(-) rarr Sn^(2+),,E^(2)=+0.15" volt"):}`
On the basis of the above data, answer the following questions :
(a) Whether `Fe^(3+)` oxidises `Ce^(3+)` or not ?
(b) Whether `I_(2)` displaces chlorine from KCl ?
(c) Whether the reaction between `FeCl_(3)` and `SnCl_(2)` occurs or not ?

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) Chemical reaction,
`Fe^(3+)+Ce^(3+) rarr Ce^(4+) +Fe^(2+)`
Two half reactions,
`{:(Fe^(3+)+e rarr Fe^(2+),,"Reduction, "E^(@)=0.76" volt"),(Ce^(3+) rarr Ce^(4+)+e^(-),,"Oxidation, "E_(o x)^(@)=-1.60" volt"),(,bar("Adding, emf"=-0.84" volt")):}`
Since, emf is neagtive the reaction does not occur, i.e., `Fe^(3+)` does not oxidise `Ce^(3+)`
(b) Chemical reaction,
`I_(2)+2KCl=2KI+Cl_(2)`
Half reactions,
`{:(I_(2)+2e^(-) rarr 2I^(-),,"Reduction, "E^(@)=0.54" volt"),(2Cl^(-) rarr Cl_(2)+2e^(-),,"Oxidation, "E_(o x)^(@)=-1.36" volt"),(,bar("Adding, emf"=-0.82" volt")):}`
since, emf is negative, the reaction does not occur, i.e., `I_(2)` does not displace `Cl_(2)` from `KCl`.
(c) Chemical reaction,
`SnCl_(2)+2FeCl_(3) rarr SnCl_(4)+2FeCl_(2)`
Half reactions,
`{:(Fe^(3+)+e^(-) rarr Fe^(2+),,"Reduction, "E^(@)=0.76" volt"),(Sn^(2+)rarrSn^(4+)+2e^(-),,"Oxidation, "E^(@)=-0.15" volt"),(,"Adding, emf"=+0.61" volt"):}`
Since, emf is positive, the reaction will occur.