Problem 2 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following is not a redox reaction? (a) \(\mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}+\mathrm{CO}_{2}\) (b) \(\mathrm{O}_{2}+2 \mathrm{H}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{Na}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaOH}+1 / 2 \mathrm{H}_{2}\) (d) \(\mathrm{SnCl}_{4} \longrightarrow \mathrm{SnCl}_{2}+\mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

Option (a) \(\mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}+\mathrm{CO}_{2}\) is not a redox reaction.

Step 1: Understand Redox Reactions

Redox reactions involve the transfer of electrons between two substances. This means one species will lose electrons (oxidation) and another will gain electrons (reduction). Identifying changes in oxidation states of elements in the reaction is key to identifying redox reactions.

Step 2: Analyze Option (a)

Look at the reaction \(\mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}+\mathrm{CO}_{2}\).Calculate oxidation states:- In \(\mathrm{MgCO}_3\), Mg is +2; C is +4; O is -2.- In \(\mathrm{MgO}\), Mg is +2; O is -2.- In \(\mathrm{CO}_2\), C is +4; O is -2.There are no changes in oxidation states of any element. Hence, this is not a redox reaction.

Step 3: Analyze Option (b)

Look at the reaction \(\mathrm{O}_{2}+2 \mathrm{H}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\).Calculate oxidation states:- \(\mathrm{O}_2\): O is 0; \(\mathrm{H}_2\): H is 0.- In \(\mathrm{H}_2\mathrm{O}\): O is -2; H is +1.Oxygen gains electrons (reduction) and hydrogen loses electrons (oxidation), making this a redox reaction.

Step 4: Analyze Option (c)

Look at the reaction \(\mathrm{Na}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{NaOH}+\frac{1}{2}\mathrm{H}_{2}\).Calculate oxidation states:- In reactants: Na is 0; H is +1 in water; O is -2.- In \(\mathrm{NaOH}\): Na is +1; H is +1; O is -2.- In \(\mathrm{H}_2\): H is 0.Sodium is oxidized, and hydrogen is reduced, making this a redox reaction.

Step 5: Analyze Option (d)

Look at the reaction \(\mathrm{SnCl}_{4} \longrightarrow \mathrm{SnCl}_{2}+\mathrm{Cl}_{2}\).Calculate oxidation states:- In \(\mathrm{SnCl}_4\): Sn is +4; Cl is -1.- In \(\mathrm{SnCl}_2\): Sn is +2; Cl is -1.- Cl changes from -1 in \(\mathrm{SnCl}_4\) to 0 in \(\mathrm{Cl}_2\), indicating oxidation.Since tin is reduced and chlorine is oxidized, this is a redox reaction.

Key Concepts

Oxidation States

Oxidation states are a way to keep track of electrons in a chemical reaction. They help us determine whether a reaction involves a transfer of electrons, which is key in redox reactions. Understanding how oxidation states work can tell us whether oxidation or reduction has occurred.

• An element's oxidation state can be thought of as its charge if all bonds were 100% ionic.
• In most compounds, the sum of the oxidation states equals the overall charge of the compound.
• For example, in \(\mathrm{MgCO}_3\), magnesium (Mg) has an oxidation state of +2, since it's typically found as a +2 ion. Carbon (C) is +4, and oxygen (O) is -2, which are also common oxidation states for these elements in compounds.
• Adding these together, \((+2 + 4*(+4) + 3*(-2)) = 0\), verifies that the compound is neutral overall.

Knowing oxidation states allows us to quickly identify redox reactions. A change in oxidation state indicates that an electron transfer has taken place, which is characteristic of a redox process.

Electron Transfer

Electron transfer is the movement of electrons from one chemical species to another. This movement is what defines a redox reaction. A change in oxidation states among the reacting species highlights this transfer.
When we analyze reactions:

• We look for changes in oxidation states to detect electron transfer.
• If an oxidation state increases, it implies the element has lost electrons—this is oxidation.
• If an oxidation state decreases, it means the element has gained electrons—this is reduction.

For example, in the reaction \(\mathrm{SnCl}_4 \longrightarrow \mathrm{SnCl}_2+\mathrm{Cl}_2\):

• Sn decreases from +4 to +2, meaning it gains electrons (reduction).
• Cl changes from -1 to 0 on forming \(\mathrm{Cl}_2\), indicating it loses electrons (oxidation).

Each electron transfer process involves both a gain and a loss of electrons, balancing out the changes in oxidation states.

Oxidation and Reduction

Oxidation and reduction are the two key processes in a redox reaction, which always occur together. When one substance loses electrons (oxidation), another gains those electrons (reduction). This interplay ensures the conservation of electric charge.
In layman's terms, you can remember this using the mnemonic OIL RIG:

• Oxidation Is Loss (of electrons)
• Reduction Is Gain (of electrons)

Let's apply this to some of the example reactions:

• In \(\mathrm{O}_2 + 2 \mathrm{H}_2 \longrightarrow 2\mathrm{H}_2\mathrm{O}\), hydrogen is oxidized from 0 to +1 as it loses electrons. Oxygen is reduced from 0 to -2 as it gains those electrons.
• In \(\mathrm{Na} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{NaOH} + \frac{1}{2}\mathrm{H}_2\), sodium becomes oxidized from 0 to +1, and hydrogen in water reduces from +1 to 0.

Recognizing how oxidation and reduction work helps in identifying the flow of electrons and following the chemical changes taking place.