Question

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E_{\ominus }\right)$ of two half cell reactions decided which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half cell reactions (acidic medium) along with their $E_{\ominus }$ (V with respect to normal hydrogen electrode) values. Using this data, obtain correct explanations for the following question
$I_2+2e^{-}\to 2I_{\ominus };E^{\ominus }=0.54$
$C{l}_2+2e^{-}\to 2C{I}_{\ominus };E^{\ominus }=1.36$
$M{n}^{3+}+e^{-}\to M{n}^{2+};E^{\ominus }=1.50$
$F{e}^{3+}+e^{-}\to F{e}^{2+};E^{\ominus }=0.77$
$O_2+4H^{⨁}+4e^{-}\to 2H_{2}I;E^{\ominus }=1.23$

While $F{e}^{3+}$ is stable, $M{n}^{3+}$ is not stable in acid solution because

• A. $O_{2}oxidizesM{n}^{2+}toM{n}^{3+}$
• B. $O_{2}oxidizesbothM{n}^{2+}toM{n}^{3+}andF{e}^{2+}toF{e}^{3+}$
• C. $F{e}^{3+}oxidizes{H}_{2}Oto{O}_2$
• D. $M{n}^{3+}oxidizes{H}_{2}Oto{O}_2$

Answer 1

The correct option is D $M{n}^{3+}oxidizes{H}_{2}Oto{O}_2$

First thing we need to realise here is that this is a simple question, just has a couple of steps you need to follow and you will be good.
What are the steps? Read the question carefully. Two reactions are being spoken about when you compare the stability of the iron (III) ion over the manganese (III) ion. These are the reactions with water the produce protons!
So, if the ions react with water, clearly they are not stable! Stable ions would like to stay in solution and not react. Let's see if manganese (III) ion reacts spontaneously with water.

$[M{n}^{3+}+e^{-}\to M{n}^{2+}]\times 4$
$2H_{2}O\to H^{⨁}+O_2+4e^{-}$
____________________________________________
$4M{n}^{3+}+2H_{2}O\to 4M{n}^{2+}+4H^{⨁}+O_2$
____________________________________________
$(E_{cell}^{\ominus }is+ve)$ which means that this reaction is spontaneous and it explains the instability of the ion in question.