IIT-JEE 2007 · Paper 2 · Chemistry Q37

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ChemistrySingle correct+4 / -1Question 37 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.)

While $\ce{Fe^{3+}}$ is stable, $\ce{Mn^{3+}}$ is not stable in acid solution because

\ce{O2}

oxidises

\ce{Mn^{2+}}

\ce{Mn^{3+}}

\ce{O2}

oxidises both

\ce{Mn^{2+}}

\ce{Mn^{3+}}

and

\ce{Fe^{2+}}

\ce{Fe^{3+}}

\ce{Fe^{3+}}

oxidises

\ce{H2O}

\ce{O2}

\ce{Mn^{3+}}

oxidises

\ce{H2O}

\ce{O2}

Answer and worked solution

Official answer: D

$\ce{Mn^{3+}}$ vs water. $E^\circ(\ce{Mn^{3+}/Mn^{2+}})=1.50$ V exceeds $E^\circ(\ce{O2/H2O})=1.23$ V, so
$\ce{4Mn^{3+} + 2H2O -> 4Mn^{2+} + O2 + 4H+}$
is spontaneous ( $E^\circ=+0.27$ V): $\ce{Mn^{3+}}$ is not stable in acid solution.

$\ce{Fe^{3+}}$ (0.77 V) cannot oxidise water, so it is stable.

(D)

Topics:Galvanic cells and EMF Transition element properties

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IIT-JEE 2007 · Paper 2, Chemistry question 37