Question

Reaction which involve redox change is/are

• A. Cu$^{+2}$ + $\bar{CN}_{(Excess)}$ $\longrightarrow$
• B. FeS $\xrightarrow{Roasting}$
• C. CO$_2$ + MnO$_4^-$ $\xrightarrow{weak alkaline solution}$
• D. Na$_2$CrO$_4$ + H$_2$SO$_4$ $\longrightarrow$

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

Let's analyze each reaction to determine if it involves a redox change (change in oxidation states).

(A) Cu$^{+2}$ + $\bar{CN}_{(Excess)}$ $\longrightarrow$

The reaction of Cu$^{2+}$ with excess cyanide involves the reduction of Cu$^{2+}$ to Cu$^+$ and the oxidation of cyanide to cyanogen (CN)$_2$. The balanced reaction is:

$2Cu^{2+} + 8CN^{-} \longrightarrow 2[Cu(CN)_4]^{3-} + (CN)_2$

Let's determine the oxidation states:

Reactants:

• Cu$^{2+}$: Oxidation state of Cu is +2.
• CN$^{-}$: Let oxidation state of C be $x$ and N be $y$. $x+y = -1$. Using electronegativity (N > C), we can assign typical oxidation states. If N is -3, C is +2. So, C is +2, N is -3.

Products:

• $[Cu(CN)_4]^{3-}$: Let oxidation state of Cu be $z$. The cyanide ligand has a charge of -1. $z + 4(-1) = -3 \implies z = +1$. Oxidation state of Cu is +1.
• $(CN)_2$: Structure is N$\equiv$C-C$\equiv$N. The C-C bond is non-polar. In the C$\equiv$N bond, N is more electronegative. Assigning oxidation states based on electronegativity, N is -3 and C is +3.

Changes in oxidation states:

• Cu: +2 $\longrightarrow$ +1 (Reduced)
• C: +2 (in CN$^{-}$) $\longrightarrow$ +3 (in (CN)$_2$) (Oxidized)
• N: -3 (in CN$^{-}$) $\longrightarrow$ -3 (in (CN)$_2$) (No change)

Since the oxidation states of Cu and C change, this is a redox reaction.

(B) FeS $\xrightarrow{Roasting}$

Roasting involves heating a sulfide ore in the presence of oxygen. A typical reaction is:

$2FeS(s) + 3.5O_2(g) \longrightarrow Fe_2O_3(s) + 2SO_2(g)$

Let's determine the oxidation states:

Reactants:

• FeS: Fe is typically +2 in FeS, S is -2.
• O$_2$: Oxidation state of O is 0.

Products:

• Fe$_2$O$_3$: O is -2. $2 \times (\text{O.S. of Fe}) + 3 \times (-2) = 0 \implies \text{O.S. of Fe} = +3$.
• SO$_2$: O is -2. $\text{O.S. of S} + 2 \times (-2) = 0 \implies \text{O.S. of S} = +4$.

Changes in oxidation states:

• Fe: +2 $\longrightarrow$ +3 (Oxidized)
• S: -2 $\longrightarrow$ +4 (Oxidized)
• O: 0 $\longrightarrow$ -2 (Reduced)

Since the oxidation states of Fe, S, and O change, this is a redox reaction.

(C) CO$_2$ + MnO$_4^-$ $\xrightarrow{weak alkaline solution}$

In this reaction, MnO$_4^-$ acts as an oxidizing agent. In weak alkaline solution, MnO$_4^-$ is typically reduced to MnO$_2$.

Let's determine the oxidation states of Mn:

• MnO$_4^-$: O is -2. $\text{O.S. of Mn} + 4 \times (-2) = -1 \implies \text{O.S. of Mn} = +7$.
• MnO$_2$: O is -2. $\text{O.S. of Mn} + 2 \times (-2) = 0 \implies \text{O.S. of Mn} = +4$.

Mn is reduced from +7 to +4.

For this to be a redox reaction, something must be oxidized. The other reactant listed is CO$_2$.

In CO$_2$: O is -2. $\text{O.S. of C} + 2 \times (-2) = 0 \implies \text{O.S. of C} = +4$.

Carbon is in its maximum possible oxidation state (+4). Therefore, CO$_2$ cannot be oxidized further.

Since MnO$_4^-$ is reduced, a reducing agent must be present. CO$_2$ cannot act as a reducing agent when carbon is in the +4 state. A reaction between only CO$_2$ and MnO$_4^-$ as listed, where one is oxidized and the other is reduced, is not possible. The listed species would not undergo a redox reaction with each other.

(D) Na$_2$CrO$_4$ + H$_2$SO$_4$ $\longrightarrow$

Sodium chromate (Na$_2$CrO$_4$) reacts with sulfuric acid (H$_2$SO$_4$) to form sodium dichromate (Na$_2$Cr$_2$O$_7$), sodium sulfate (Na$_2$SO$_4$), and water. This is an interconversion between chromate and dichromate ions in acidic medium. The reaction is:

$2Na_2CrO_4(aq) + H_2SO_4(aq) \longrightarrow Na_2Cr_2O_7(aq) + Na_2SO_4(aq) + H_2O(l)$

Let's determine the oxidation states:

Reactants:

• Na$_2$CrO$_4$: Na is +1, O is -2. $\text{O.S. of Cr} + 4 \times (-2) = -2$ (for CrO$_4^{2-}$ ion) $\implies \text{O.S. of Cr} = +6$.
• H$_2$SO$_4$: H is +1, O is -2, S is +6.

Products:

• Na$_2$Cr$_2$O$_7$: Na is +1, O is -2. $2 \times (\text{O.S. of Cr}) + 7 \times (-2) = -2$ (for Cr$_2$O$_7^{2-}$ ion) $\implies 2 \times \text{O.S. of Cr} = +12 \implies \text{O.S. of Cr} = +6$.
• Na$_2$SO$_4$: Na is +1, O is -2, S is +6.
• H$_2$O: H is +1, O is -2.

Changes in oxidation states:

• Cr: +6 $\longrightarrow$ +6 (No change)
• Na: +1 $\longrightarrow$ +1 (No change)
• S: +6 $\longrightarrow$ +6 (No change)
• O: -2 $\longrightarrow$ -2 (No change)
• H: +1 $\longrightarrow$ +1 (No change)

Since no element changes its oxidation state, this is not a redox reaction. It is an acid-base reaction (H$^+$ reacting with the base CrO$_4^{2-}$ to form the conjugate acid Cr$_2$O$_7^{2-}$) and a metathesis/double displacement reaction.

Based on the analysis, reactions (A) and (B) involve redox changes, while reactions (C) and (D) do not.