Match Column I with Column II related to the figure given below and then select the appropriate option from the codes given below.

Column I Column II
A. Figure represents 1. Reduction
B. $O_{2} (g) + 4 H^{+} (aq) + 4 e^{-} ⟶ 2 H_{2} O (l)$ 2. Corrosion of iron of atmosphere
C. $Fe (s) ⟶ {Fe}^{2 +} (aq) + 2 e^{-}$ 3. Oxidation
D. $2 {Fe}^{2 +} (aq) + 2 H_{2} O (l) + \frac{1}{2} O_{2} ⟶ {Fe}_{2} O_{3} + 4 H^{+} (aq)$

Moderate

Solution

Given figure represents the corrosion of iron in the atmosphere. Oxidation occurs at anode and reduction occurs at cathode
Cathode = $\begin{array}{l} O_{2} (g) + 4 H^{+} (aq) + 4 e^{- -} ⟶ 2 H_{2} O (l); \\ E_{H^{+} |O_{2}| H_{2} O}^{\circ} = 1 .23 V \end{array}$
Anode = $2 Fe (s) ⟶ 2 {Fe}^{2 +} (aq) + 4 e^{-}; E_{{Fe}^{2 +} / Fe}^{\circ} = - 0 .44 V$
Overall reaction = $\begin{array}{l} 2 Fe (s) + O_{2} (g) + 4 H^{+} (aq) ⟶ 2 {Fe}^{2 +} (aq) \\ + 2 H_{2} O (l); E_{cell}^{\circ} = 1 .67 V \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME

	Column I		Column II
A.	Figure represents	1.	Reduction
B.	$O_{2} (g) + 4 H^{+} (aq) + 4 e^{-} ⟶ 2 H_{2} O (l)$	2.	Corrosion of iron of atmosphere
C.	$Fe (s) ⟶ {Fe}^{2 +} (aq) + 2 e^{-}$	3.	Oxidation
D.	$2 {Fe}^{2 +} (aq) + 2 H_{2} O (l) + \frac{1}{2} O_{2} ⟶ {Fe}_{2} O_{3} + 4 H^{+} (aq)$