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  • A solution contains Fe2+, Fe3+ and I- ions. This solution was treated with iodine at 35°C. Eo for Fe3+/Fe2+ is +0.77 V and

A solution contains Fe2+, Fe3+ and I- ions. This solution was treated with iodine at 35°C. Eo for Fe3+/Fe2+ is +0.77 V and Eo for \mathrm{I_2/2I^- = 0.536\ V}.The favourable redox reaction is :

Option: 1

I- will be oxidised to I2


Option: 2

Fe2+ will be oxidised to Fe3+


Option: 3

I2 will be reduced to I-


Option: 4

There will be no redox reaction


Answers (1)

For the favourable redox reaction, \mathrm{E_{cell}} Should be greater than Zero.

Due to \mathrm{E^o } of \mathrm{Fe^{3+}/Fe^{2+}} is greater than of \mathrm{I_2/I^-} So, only this redox is possible. 

\mathrm{2 I^{-} \rightarrow I_{2}+2 e^{-} \ \ \ \ \(\text{oxidation half - reaction})\ \(E_{\text {left }}^{\circ}=0.536 \mathrm{V}\)}

\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} \ \ \ \(\text{reduction half - reaction})\ \(\mathrm{E_{\text {right }}^{\circ}=0.77 \ V}\) \\-------------------------------

2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_{2}

So, For the above reaction

\ \mathrm{E_{cell}}=\mathrm{E}_{\text {right }}^{\circ}-\mathrm{E}_{\text {left }}^{\circ}\\ 

\mathrm{E_{cell}= 0.77 - 0.536}

\mathrm{E_{cell}= 0.23}

This value is positive. so, the reaction will take place.

So, I- will be oxidised to I2 and F3+ will be reduced to Fe2+.

Option 1 is correct.

Posted by

Kshitij

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