JEE Advanced 2013 · Paper 2 · Chemistry Q39

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ChemistryList match+3 / -1Question 39 of 60

The standard reduction potential data at 25 °C is given below.

$E^\circ(\ce{Fe^{3+}},\ce{Fe^{2+}})=+0.77$ V; $E^\circ(\ce{Fe^{2+}},\ce{Fe})=-0.44$ V;
$E^\circ(\ce{Cu^{2+}},\ce{Cu})=+0.34$ V; $E^\circ(\ce{Cu+},\ce{Cu})=+0.52$ V;
$E^\circ[\ce{O2(g) + 4H+ + 4e- -> 2H2O}]=+1.23$ V; $E^\circ[\ce{O2(g) + 2H2O + 4e- -> 4OH-}]=+0.40$ V;
$E^\circ(\ce{Cr^{3+}},\ce{Cr})=-0.74$ V; $E^\circ(\ce{Cr^{2+}},\ce{Cr})=-0.91$ V

Match $E^\circ$ of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists:

List I

P
$E^\circ(\ce{Fe^{3+}},\ce{Fe})$
Q
$E^\circ(\ce{4H2O <=> 4H+ + 4OH-})$
R
$E^\circ(\ce{Cu^{2+} + Cu -> 2Cu+})$
S
$E^\circ(\ce{Cr^{3+}},\ce{Cr^{2+}})$

List II

1
−0.18 V
2
−0.4 V
3
−0.04 V
4
−0.83 V

P → 4; Q → 1; R → 2; S → 3

P → 2; Q → 3; R → 4; S → 1

P → 1; Q → 2; R → 3; S → 4

P → 3; Q → 4; R → 1; S → 2

Answer and worked solution

Official answer: D

Combine via ΔG = −nFE.

P, Fe³⁺/Fe: $\dfrac{1(0.77)+2(-0.44)}3=-0.037\approx-0.04$ V. → 3
Q, $\ce{4H2O <=> 4H+ + 4OH-}$ : $E=0.40-1.23=-0.83$ V. → 4
S, Cr³⁺/Cr²⁺: $3(-0.74)-2(-0.91)=-0.40$ V. → 2
R, $\ce{Cu^{2+} + Cu -> 2Cu+}$ : $E^\circ(\ce{Cu^{2+}/Cu+})=2(0.34)-0.52=0.16$ V, so $E=0.16-0.52=-0.36$ V. This is not in List II; the remaining value, −0.18 V (half of it), is the one the key pairs with R.

P, Q and S alone fix the code 3 4 1 2.

(D)

Topics:Gibbs energy and EMF

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JEE Advanced 2013 · Paper 2, Chemistry question 39