IIT-JEE 2007 · Paper 2 · Chemistry Q36

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ChemistrySingle correct+4 / -1Question 36 of 66

Passage

Paragraph for Question Nos. 36 to 38

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential ( $E^\circ$ ) of two half-cell reactions decide 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 $E^\circ$ (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 36-38.

Half-cell reaction	$E^\circ$ (V)
$\ce{I2 + 2e- -> 2I-}$	0.54
$\ce{Cl2 + 2e- -> 2Cl-}$	1.36
$\ce{Mn^{3+} + e- -> Mn^{2+}}$	1.50
$\ce{Fe^{3+} + e- -> Fe^{2+}}$	0.77
$\ce{O2 + 4H+ + 4e- -> 2H2O}$	1.23

(The paragraph is not printed in the official archive copy (which reproduces only the questions with their answers); its text is restored from the JEE 2007 question paper.)

Among the following, identify the correct statement.

Chloride ion is oxidised by

\ce{O2}

\ce{Fe^{2+}}

is oxidised by iodine

Iodide ion is oxidised by chlorine

\ce{Mn^{2+}}

is oxidised by chlorine

Answer and worked solution

Official answer: C

Rule. A species with a higher reduction potential oxidises the reduced form of a couple with a lower one.

(A) $\ce{O2}$ (1.23) cannot oxidise $\ce{Cl-}$ (1.36).
(B) $\ce{I2}$ (0.54) cannot oxidise $\ce{Fe^{2+}}$ (0.77).
(C) $\ce{Cl2}$ (1.36) oxidises $\ce{I-}$ (0.54): $E^\circ_\text{cell}=+0.82$ V. Correct.
(D) $\ce{Cl2}$ (1.36) cannot oxidise $\ce{Mn^{2+}}$ (1.50).

(C)

Topics:Galvanic cells and EMF Oxidation number

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IIT-JEE 2007 · Paper 2, Chemistry question 36