Question

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 favourable redox reaction is

Answer 1

Answer:

Explanation:

Check pics for explanation

Answer 2

Answer: The favorable redox reaction is written below.

Explanation:

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

We are given:

$E^o_{Fe^{3+}/Fe^{2+}}=0.77V\\E^o_{I_2/2I^-}=0.536V$

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=0.77-0.536=0.234V$

Here, iron will get reduced and iodine will get oxidized.

Half reaction follows:

Oxidation half reaction:  $2I^-\rightarrow I_2+2e^-$

Reduction half reaction:  $Fe^{3+}+e^-\rightarrow Fe^{2+}$    ( ×  2)

Net redox reaction follows:

$2I^-+2Fe^{3+}\rightarrow I_2+2Fe^{2+}$

Hence, the favorable redox reaction is written above