Question

This question and the following one were a part of a linked comprehension in the (IIT-JEE, 2008) exam.

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 follwing questions.
$I_2+2e^{-}\to 2I_{\ominus };E^{\ominus }=0.54$
$C{l}_2+2e^{-}\to 2C{I}_{\ominus };E^{\ominus }=1.36$
$M{n}^{3+}+e^{-}\to M{n}^{2+};E^{\ominus }=1.50$
$F{e}^{3+}+e^{-}\to F{e}^{2+};E^{\ominus }=0.77$
$O_2+4H^{⨁}+4e^{-}\to 2H_{2}I;E^{\ominus }=1.23$

Among the following, identify the correct statement.

• A. Chloride ion is oxidized by $O_2.$
• B. $F{e}^{2+}$ is oxidized by iodine.
• C. Iodide ion is oxidized by chlorine.
• D. $M{n}^{2+}$ is oxidized by chlorine.

Answer 1

The correct option is C Iodide ion is oxidized by chlorine.

You need basic redox ideas here. We are simply applying what we learned in the electrochemical series here.
All the values given to us are reduction potential values since all half reactions have electrons on the left hand side of the reaction. If electrons are added to a species, it is reduced, this idea is straightforward.
Also, if the overall reaction is feasible, it will have a positve $E^0$ value right?
Among the following options option (c) makes sense. Let's see why:

$2I^{\ominus }+C{l}_2\to I_2+2C{l}^{\ominus };E_{cell}^{\ominus }=1.36-0.54=0.82V$
$E_{cell}^{\ominus }>0$ . Thus, the cell reaction is feasible.