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Introduction to Oxidation and Reduction

A solution co...

Question

A solution contains ${F e}^{+ 2}, {F e}^{+ 3}$ and $I^{-}$ ions. This solution was treated with iodine at $35^{o} C$ . Then the favourable redox reaction is:
(Given that $E_{{F e}^{+ 3} / {F e}^{+ 2}}^{o} = + 0.77 V; E_{I_{2} / I^{-}}^{o} = 0.536 V$ )

A

I_{2}

will be reduced to

I^{-}

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B

there will be no redox reaction

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C

I^{-}

will be oxidised to

I_{2}

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D

{F e}^{+}

will be oxidised to

{F e}^{+ 3}

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Solution

The correct option is A $I_{2}$ will be reduced to $I^{-}$
${2 I}^{-} \to I_{2} + {2 e}^{-} (o x i d a t i o n h a l f - r e a c t i o n)$
$E_{O x i d a t i o n}^{0} = - 0.536 V$
${F e}^{3 +} + e^{-} \to {F e}^{2 +} (r e d u c t i o n h a l f - r e a c t i o n)$
$E_{R e d u c t i o n}^{0} = - 0.77 V$
----------------------------------------------------------------------------------------------------------------------
${2 F e}^{3 +} + {2 I}^{-} \to {2 F e}^{2 +} + I_{2}$
$E^{0} = E_{o x i d a t i o n}^{0} + E_{R e d u c t i o n}^{0}, + v e$
Hence the reaction will take place.
${2 I}^{-} \to I_{2} + {2 e}^{-} (o x i d a t i o n h a l f - r e a c t i o n)$
$E_{O x i d a t i o n}^{0} = - 0.536 V$
${F e}^{3 +} + e^{-} \to {F e}^{2 +} (r e d u c t i o n h a l f - r e a c t i o n)$
$E_{R e d u c t i o n}^{0} = - 0.77 V$
------------------------------------------------------------------------------------------------------------------------
$2 F e^{3 +} + {2 I}^{-} \to {2 F e}^{2 +} + I_{2};$
$E^{0} = E_{o x i d a t i o n}^{0} + E_{r e d u c t i o n}^{0}; + v e$

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Similar questions

Q. A solution contains

{F e}^{2 +}, {F e}^{3 +}

I^{-}

ions. This solution was treated with iodine at

35^{o} C

E^{0}

{F e}^{3 +} / {F e}^{2 +}

+ 0.77 V

E^{o}

I_{2} / 2 I^{-} = 0.536 V

. The favourable redox reaction is:

Q. If equal volumes of 0.1

M K M n O_{4}

M K_{2} C r_{2} O_{7}

solutions are allowed to oxidise

F e^{2 +}

F e^{3 +}

in acidic medium, then

F e^{2 +}

oxidised will be:

F e

F e^{2 +}

by reaction with

1.0 M H C l

E^{\circ}

F e / F e^{2 +} = + 0.44 v o l t

.

Q. If equal volumes of

0.1 M K M n O_{4}

0.1 M K_{2} C r_{2} O_{7}

solutions are allowed to oxidise

F e^{2 +} t o F e^{3 +}

in acidic medium, then

F e^{2 +}

oxidised will be :

F e_{(s)}

F e^{2 +}

by the reaction with

1 M H C l

E^{\circ}

F e / F e^{2 +} = + 0.44 V

)

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