A solution contains Fe2- Fe3- and I- ions- This solution was treated with iodine at 35-C- E0 for Fe3- -Fe2- is -0-77 V and E0 for I2-2I- - 0-536 V- The favourable redox reaction is -AIPMT MAINS 2011

The favorable reaction is the oxidation of iodide ion to iodine. nbsp; The couple which has higher reduction potential will be reduce and couple with lower re ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Representation of Cell

A solution co...

Question

A solution contains $F e^{2 +}, F e^{3 +} a n d I^{-} i o n s .$ This solution was treated with iodine at $35^{\circ} C . E^{0}$ for $F e^{3 +} / F e^{2 +} i s + 0.77 V a n d E^{0} f o r I_{2} / 2 I^{-} = 0.536 V .$ The favourable redox reaction is -
(  AIPMT MAINS 2011)

F e^{2 +} will be oxidized to F e^{3 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{2} will be the reduced to I^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I^{-} will be oxidized to I_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

There will be no redox reaction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $I^{-} will be oxidized to I_{2}$
The favorable reaction is the oxidation of iodide ion to iodine. The couple which has higher reduction potential will be reduce and couple with lower reduction potential gets oxidised. Here, $F e^{3 +} / F e^{2 +}$ couple has higher reduction potential so $F e^{3 +}$ get reduced to $F e^{2 +}$ . Therefore, $I^{-}$ get oxidised to $I_{2}$ . $\begin{matrix} e^{-} + F e^{+ 3} \to F e^{2 +}; E = 0.77 V 2 I^{-} \to I_{2} + 2 e^{-}; E = - 0.536 V 2 F e^{+ 3} + 2 I^{-} \to 2 F e^{+ 2} + I_{2}; E = E_{r e d} + E_{o x} E = 0.77 - 0.536 = 0.234 V \end{matrix}$ Since the value of the cell potential is positive, the reaction will take place, Hence, option C is correct.

Suggest Corrections

Similar questions

Q. A solution contains

F e^{2 +}, F e^{3 +} a n d I^{-} i o n s .

This solution was treated with iodine at

35^{\circ} C . E^{0}

for

F e^{3 +} / F e^{2 +} i s + 0.77 V a n d E^{0} f o r I_{2} / 2 I^{-} = 0.536 V .

The favourable redox reaction is -

Q. A solution contains

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

and

I^{-}

ions. This solution was treated with iodine at

35^{o} C

E^{0}

for

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

+ 0.77 V

and

E^{o}

for

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

. The favourable redox reaction is:

Q. 3A solution contains Fe2, Fe3 and I ions. This solutionwas treated with iodine at 35^° C. E^° for Fe3 /Fe2+ is+0.77 V and E^° for I/21 =0.536 V. The favourableredox reaction is\lbrack AIPMT (Mains)-2011\rbrack

A solution contains Fe2+, Fe3+ and I ions. This solution was treated with iodine at 35C. E for Fe3+/Fe2+ is +0.77 V and E for I2/2I = 0.536 V.The favorable redox reaction is:

(1) $I^{-}$ will be oxidized to $I_{2}$
(2) $F e^{2 +}$ will be reduced to $F e^{3 +}$
(3) $I_{2}$ will be reduced to $I^{-}$
(4) there will be no redox reaction takes place

Please explain

Q. What will be the

E^{0}

value for the given half cell reaction?

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

Given:

E^{0}

value for the half cell reaction,

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

Join BYJU'S Learning Program

A solution contains Fe2+,Fe3+ and I− ions. This solution was treated with iodine at 35∘C. E0 for Fe3+/Fe2+ is +0.77 V and E0 for I2/2I−=0.536 V. The favourable redox reaction is - (  AIPMT MAINS 2011)