Electrochemistry Result Question 15
15. The standard reduction potential data at $25^{\circ} \mathrm{C}$ is given below.
$ \begin{aligned} & E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{V} ; E^{\circ}\left(\mathrm{Fe}^2 / \mathrm{Fe}\right)=-0.44 \mathrm{V} \\ & E^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)=+0.34 \mathrm{V} ; E^{\circ}\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)=+0.52 \mathrm{V} \\ & \left.E^{\circ}\left(\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 e^{-}\right) \longrightarrow 2 \mathrm{H}_2 \mathrm{O}\right)=+1.23 \mathrm{V} \\ & \left.E^{\circ}\left(\mathrm{O}_2(g)+2 \mathrm{H}_2 \mathrm{O}+4 e^{-}\right) \longrightarrow 4 \mathrm{OH}\right)=+0.40 \mathrm{V} \\ & 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 rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.
(2013 Adv.)
| Column I | Column II | ||
|---|---|---|---|
| P. | $E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}\right)$ | 1. | -0.18 V |
| Q. | $E^{\circ}\left(4 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 4 \mathrm{H}^{+}+4 \mathrm{OH}^{-}\right)$ | 2. | -0.4 V |
| R. | $E^{\circ}\left(\mathrm{Cu}^{2+}+\mathrm{Cu} \longrightarrow 2 \mathrm{Cu}^{+}\right)$ | 3. | -0.04 V |
| S. | $E^{\circ}\left(\mathrm{Cr}^{3+}, \mathrm{Cr}^{2+}\right)$ | 4. | -0.83 V |
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Answer:
Correct Answer: 15. ( d )
Solution:
BETA
