Grade 11 : Chemistry - Redox Reactions (Set 4)

Explanation: A lower (more negative) standard reduction potential indicates a greater tendency to undergo oxidation, making the metal a stronger reducing agent. Zinc has the most negative potential (-0.76 V), so $\text{Zn}$ is the strongest reducing agent among the given species.

Explanation: Let $x$ be the oxidation state of $\text{Cr}$. Potassium is +1 and Oxygen is -2.
$$2(+1) + 2(x) + 7(-2) = 0$$
$$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$$
Thus, the oxidation state of chromium is +6.

Explanation: $\text{Fe}^{2+}$ is converted to $\text{Fe}^{3+}$, losing one electron (oxidation state increases from +2 to +3). Since $\text{Fe}^{2+}$ undergoes oxidation, it acts as the reducing agent.

Explanation: Oxidation half-reaction: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$ (multiply by 5).
Reduction half-reaction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (multiply by 2).
Combining gives: $2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$.
So $a=2, b=5, c=16$.

Explanation: In $\text{Na}_2\text{S}_2\text{O}_3$, sodium is +1 and oxygen is -2.
$$2(+1) + 2(x) + 3(-2) = 0 \implies 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2$$
The average oxidation state of sulfur is +2.

Explanation: In $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$, oxygen in $\text{H}_2\text{O}_2$ has an oxidation number of -1. It is reduced to -2 in $\text{H}_2\text{O}$ and oxidized to 0 in $\text{O}_2$. The same element is simultaneously oxidized and reduced.

Explanation: $\text{H}_2\text{SO}_5$ has structure $\text{HO}-\text{S}(=\text{O})_2-\text{O}-\text{O}-\text{H}$. Two oxygen atoms belong to a peroxide linkage (oxidation state -1) and three are oxide linkages (-1 each for $\text{OH}$ and $\text{O}$).
$$2(+1) + x + 3(-2) + 2(-1) = 0 \implies x - 6 = 0 \implies x = +6$$
Sulfur cannot exceed its maximum valence group state of +6.

Explanation: In acidic medium, $\text{MnO}_4^-$ gains $5e^-$ ($\text{Mn}^{+7} \rightarrow \text{Mn}^{+2}$), while $\text{Fe}^{2+}$ loses $1e^-$ ($\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$).
Equating electron change: $5 \times n(\text{KMnO}_4) = 1 \times n(\text{FeSO}_4)$.
For 1 mole of $\text{FeSO}_4$, $n(\text{KMnO}_4) = \frac{1}{5} = 0.2\text{ mole}$.

Explanation: $\text{Fe}_3\text{O}_4$ is a stoichiometric mixture of iron(II) oxide and iron(III) oxide, formulated as $\text{FeO} \cdot \text{Fe}_2\text{O}_3$. Thus, it contains one $\text{Fe}^{2+}$ ion (oxidation state +2) and two $\text{Fe}^{3+}$ ions (oxidation state +3).

Explanation: Statement I is correct: oxidation is defined as loss of electrons leading to an increased oxidation state. Statement II is incorrect because a reducing agent loses electrons and undergoes oxidation, causing another species to be reduced.

Explanation: Oxidation always occurs at the anode. In a galvanic Daniell cell, zinc has a lower reduction potential than copper, so zinc undergoes oxidation ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$) at the anode (negative terminal).

Explanation: Elemental iodine has oxidation state 0. In $\text{NaI}$, iodine is -1 (reduced). In $\text{NaIO}_3$, iodine is +5 (oxidised). Because iodine undergoes oxidation and reduction simultaneously, it is both oxidised and reduced.

Explanation: In acidic medium, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$. The change in oxidation number of manganese per formula unit is $+7 - (+2) = 5$. Equivalent mass = $\frac{\text{Molar Mass}}{n\text{-factor}} = \frac{M}{5}$.

Explanation: $\text{NH}_4\text{NO}_3$ dissociates into $\text{NH}_4^+$ and $\text{NO}_3^-$.
For $\text{NH}_4^+$: $x + 4(+1) = +1 \implies x = -3$.
For $\text{NO}_3^-$: $y + 3(-2) = -1 \implies y = +5$.
Hence, the oxidation states are -3 and +5.

Explanation: Thermal decomposition of calcium carbonate: $\text{Ca}$ remains +2, $\text{C}$ remains +4, and $\text{O}$ remains -2 in reactant and products. Since no elements change oxidation states, it is a non-redox decomposition reaction.

Explanation: In $\text{IO}_3^-$, iodine has an oxidation number of +5. In $\text{I}_2$, iodine has an oxidation state of 0. Per iodine atom, the change is $5 - 0 = 5$ units decrease. For 1 mole of $\text{I}_2$ (containing 2 moles of iodine atoms), total change $= 2 \times 5 = 10$ units decrease.

Explanation: To oxidize $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ ($E^\circ = +0.77\text{ V}$), the halogen couple must have a higher standard reduction potential. $\text{Cl}_2$ (+1.36 V) and $\text{Br}_2$ (+1.09 V) both have higher $E^\circ$ values than $\text{Fe}^{3+}/\text{Fe}^{2+}$, whereas $\text{I}_2$ (+0.54 V) has a lower value. Thus, only $\text{Br}_2$ and $\text{Cl}_2$ can oxidize $\text{Fe}^{2+}$.

Explanation: The linear structure is $\text{O}=\text{C}=\text{C}=\text{C}=\text{O}$. Each terminal carbon is bonded to an electronegative oxygen atom via a double bond, giving them an oxidation state of +2. The central carbon is bonded only to two identical carbon atoms, giving it an oxidation state of 0. Thus, the oxidation states are $+2, 0, +2$.

Explanation: Balancing in basic medium:
$ ext{MnO}_4^- + 2 ext{H}_2 ext{O} + 3e^- ightarrow ext{MnO}_2 + 4 ext{OH}^-$.
Oxygen and charge balance require $4 ext{OH}^-$ on the right side of the balanced equation.

Explanation: From the balanced equation, 1 mole of dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) reacts stoichiometrically with 3 moles of hydrogen sulphide ($\text{H}_2\text{S}$). Thus, 1 mole of $\text{K}_2\text{Cr}_2\text{O}_7$ oxidizes 3 moles of $\text{H}_2\text{S}$.