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  4. Redox Reactions Chapter-Wise Test 14

Redox Reactions Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the oxidation number of iodine in \( \ce{I2} \)?

In its elemental form, \( \ce{I2} \), iodine has an oxidation number of 0.

-1
+1
+2
0
4

In the reaction \( \ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O} \), which species is the oxidizing agent?

\( \ce{HNO3} \) (N: +5) is reduced to \( \ce{NH4NO3} \) (N: -3) and oxidizes Zn (0 to +2).

\( \ce{Zn} \)
\( \ce{Zn(NO3)2} \)
\( \ce{HNO3} \)
\( \ce{NH4NO3} \)
3

In the reaction \( \ce{2S + 3O2 -> 2SO3} \), how many electrons are transferred per sulfur atom?

S (0) becomes +6 in \( \ce{SO3} \). Change = +6 - 0 = 6 electrons per S atom.

2
4
8
6
4

Which reaction involves a redox process where the oxidizing agent contains nitrogen?

In \( \ce{3Zn + 2NO -> 3ZnO + N2} \), \( \ce{NO} \) (N: +2) is reduced to \( \ce{N2} \) (N: 0), oxidizing Zn (0 to +2).

\( \ce{2Na + Cl2 -> 2NaCl} \)
\( \ce{3Zn + 2NO -> 3ZnO + N2} \)
\( \ce{Fe + S -> FeS} \)
\( \ce{2H2O2 -> 2H2O + O2} \)
2

In the reaction \( \ce{2NaCl + F2 -> 2NaF + Cl2} \), which species is reduced?

F (0 in \( \ce{F2} \)) becomes -1 in \( \ce{NaF} \), gaining electrons.

\( \ce{NaCl} \)
\( \ce{Cl2} \)
\( \ce{NaF} \)
\( \ce{F2} \)
4

What is the oxidation number of oxygen in \( \ce{KO2} \)?

In superoxides, O has an oxidation number of -1/2. For \( \ce{KO2} \): K = +1, so \( +1 + 2x = 0 \), \( 2x = -1 \), \( x = -1/2 \).

-1
+2
-2
-1/2
4

In the reaction \( \ce{2Na + S -> Na2S} \), which species is the oxidizing agent?

The oxidizing agent is reduced. S (0) gains electrons to become S (-2) in \( \ce{Na2S} \), oxidizing Na (0 to +1).

\( \ce{S} \)
\( \ce{Na} \)
\( \ce{Na2S} \)
None of these
1

What is the oxidation number of tellurium in \( \ce{TeO3^2-} \)?

Let Te = \( x \). O = -2. Equation: \( x + 3(-2) = -2 \), \( x - 6 = -2 \), \( x = +4 \).

+2
+6
+4
+3
3

Which species is oxidized in the reaction \( \ce{2FeCl3 + H2 -> 2FeCl2 + 2HCl} \)?

Oxidation is loss of electrons. \( \ce{H2} \) (0) becomes \( \ce{H^+} \) (+1) in \( \ce{HCl} \), losing electrons.

\( \ce{FeCl3} \)
\( \ce{H2} \)
\( \ce{FeCl2} \)
\( \ce{HCl} \)
2

What is the oxidation number of iron in \( \ce{Fe3(PO4)2} \)?

Let Fe = \( x \). \( \ce{PO4^3-} \) has P (+5) and O (-2): \( +5 + 4(-2) = -3 \). For \( \ce{Fe3(PO4)2} \): \( 3x + 2(-3) = 0 \), \( 3x - 6 = 0 \), \( x = +2 \).

+2
+3
+6
+4
1

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