Question

The standard reduction potential data at 25 °C is given below.
$$
\begin{aligned}
& \mathrm{E}^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V} ; \
& \mathrm{E}^{\circ}\left(\mathrm{Fe}^{2+}, \mathrm{Fe}\right)=-0.44 \mathrm{~V} \
& \mathrm{E}^{\circ}\left(\mathrm{Cu}^{2+}, \mathrm{Cu}\right)=+0.34 \mathrm{~V} ; \
& \mathrm{E}^{\circ}\left(\mathrm{Cu}^{+}, \mathrm{Cu}\right)=+0.52 \mathrm{~V} \
& \mathrm{E}^{\circ}\left[\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\right]=+1.23 \mathrm{~V} ; \
& \mathrm{E}^{\circ}\left[\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}+4 e^{-} \rightarrow 4 \mathrm{OH}^{-}\right]=+0.40 \mathrm{~V} \
& \mathrm{E}^{\circ}\left(\mathrm{Cr}^{3+}, \mathrm{Cr}\right)=-0.74 \mathrm{~V} \
& E^{\circ}\left(\mathrm{Cr}^{2+}, \mathrm{Cr}\right)=-0.91 \mathrm{~V}
\end{aligned}
$$
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. | $E^{\circ}\left(Fe^{3+}, Fe\right)$ |
| Q. | $E^{\circ}\left(4H_{2}O \rightleftharpoons 4H^{+}+4OH^{-}\right)$ |
| R. | $E^{\circ}\left(\mathrm{Cu}^{2+}+\mathrm{Cu} \rightarrow 2\mathrm{Cu}^{+}\right)$ |
| S. | $E^{\circ}\left(\mathrm{Cr}^{3+}, \mathrm{Cr}^{2+}\right)$ |

| List II | |
| --- | --- |
| 1. | $-0.18 V$ |
| 2. | -0.4 V |
| 3. | $-0.04 V$ |
| 4. | $-0.83 V$ |

Accepted Answer

(D) P-3, Q-4, R-1, S-2

Answer

(A) P-4, Q-1, R-2, S-3

Answer

(B) P-2, Q-3, R-4, S-1

Answer

(C) P-1, Q-2, R-3, S-4

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Redox Reactions and Electrochemistry

Solution

More JEE Advanced questions on Redox Reactions and Electrochemistry

List I
P.	$E^{\circ} (F e^{3 +}, F e)$
Q.	$E^{\circ} (4 H_{2} O ⇌ 4 H^{+} + 4 O H^{-})$
R.	$E^{\circ} (Cu^{2 +} + Cu \to 2 Cu^{+})$
S.	$E^{\circ} (Cr^{3 +}, Cr^{2 +})$

List II
1.	$- 0.18 V$
2.	-0.4 V
3.	$- 0.04 V$
4.	$- 0.83 V$