Explore topic-wise InterviewSolutions in .

Correct option is: (A) positive

Metals are electropositive in nature. Therefore, oxidation number of metal ions is always positive.

Correct option is: (A) positive

Only those electrolytes for which cations and anions have nearly the same ionic mobilities (i.e. distance traveled by an ion per second under a potential gradient of one voltmeter) are used as electrolytes in the salt bridge. Thus KCl, KNO₃, K₄SO₄, and NH₄NO₃ are preferred over NaCl, NaNO₃ and Na₂SO₄.

Among cations, H⁺ ion has the highest ionic mobility and among anions, OH^- has the highest ionic mobility.

The oxidation number of alkali metals in their compounds is +1.

+1 is the oxidation state of K in KO₂.

Let the oxidation number of Cr be x

Sum of the oxidation number of various atoms in K₂Cr₂O₇

= 2 × (+1) + 2x + 7 × (– 2) = 0

= 2 + 2x – 14 = 2x – 12 = 0

x = \(\frac{12}{2}\) = +6

\(\therefore\)Oxidation number of Cr in K₂CrO₇ = +6

MnO₄^- + I^- → MnO₂ + IO₃^- (alkaline medium)

H₂O + 2MnO₄^- + I^- → 2MnO₂ + IO₃^- + 2OH^- is balanced equation.

HA₄Cl₄ 1(ON of H) + 1C (ON of A₄) + (ON of Cl) = 0 

(+1) + x + 4(-1) = 0 

=> 1 + x - 4 = 0 

=> x = +3

NaClO₃ : 1(+1) + (oxidation no. of Cl) +3 (-2) = 0 

oxidation no. of Cl = +6 – 1 = +5

KMnO₄ : 1(+1) + (oxidation no. of Mn) + 4(-2) = 0 

+1 + (oxidation no. of Mn) – 8 = 0

∴ oxidation no. of Mn = +7

Reducing agent is a substance which can lose electrons easily. Li is the best reducing agent.

It increases.

Step-1: First we write the skeleton ionic equation, equation

\(\overset{+7}MnO_4^-(aq)+2\overset{-1}{I}(aq)\rightarrow \)  \(M\overset{+4}{nO_2}(s) +\overset{o}{I_2}(s)+O_2(s)\)

Step-2: The two half reaction are: Oxidation half reaction

\(\overset{-1}{I}(aq)\) → \(+\overset{o}{I_2}(s)\)

Reduction half reaction,

\(\overset{+7}{MnO_4^-}(aq)\rightarrow Mn\overset{+4}{O_2}(s)\)

Step-3: Balance the iodine atoms in oxidation half reaction

2I (aq) → I₂ (s)

Step-4: To balance O-atoms in oxidation half reaction, we add two water molecules to the right side.

MnO₄^-(aq) → MnO₂(s) + 2H₂O(l)

To balance the H-atoms, we add 4H⁺ ions to the left side.

MnO₄^-(aq) + 4H⁺ → MnO₂(s) + 2H₂O(l)

As the reaction takes place in basic medium, so for four H⁺ ions, we add 4 OH^– ions to both side of the equation.

MnO₄^-(aq) + 4H⁺ + 4OH^-(aq)  → MnO₂(s) 2H₂O(l) + 4OH^-(aq)

Replacing the H⁺ and OH^– ions with H₂O, then equation becomes:

MnO₄^-(aq) + 4H₂O(l)  → MnO₂(s) + 2H₂O(l) + 4OH^-(aq)

or

MnO₄^-(aq) + 2H₂O(l) → MnO₂(s) + 4OH^-(aq)

Step-5: Balance the changes of the two-half reaction

\(\overset{-1}{2I^-}(aq)\) → \(\overset{o}{I_2}(s)\) + 2e^-

\(\overset{+7}{MnO_4}(a) \) + 2H₂O (I) + 3e^- → \(\overset{+4}{MnO_2}(s) \) + 4OH^-(aq)

Now, to equalize the number of electrons, we multiply the oxidation half reaction by 3 and the reduction half reaction by 2, then, we get

6I^-(aq) → 3I₂(s) + 6e^-

2MnO₄^- (aq) + 4H₂O(I) + 6e^- → 132 MnO₂(s) + 8OH^-(aq)

Step-6: Add two half reactions to obtain the net reaction after cancelling the electrons on both sides. The net equation is:

6I^-(aq) + 2MnO₄^- (aq) + 4H₂O(l) → 3I₂ (s) + 2MnO₂(s) + 8OH^-(aq)

A final verification shows that the equation is balanced in respect of the number of atoms and charges in both sides.

(i) Standard reduction potential of A

i.e. E^Θ _A+/A = – 0.76 V

Standard reduction potential of B

i.e. E^Θ _Bn+/B = + .34 V

The electrode having more negative value of E^Θ acts as anode and the other one acts as cathode. Hence metal B acts as cathode as cathode. Hence, metal B acts as cathode and A acts as anode.

(ii) The flow of electrons taken place from anode to cathode.

Zn²⁺ is strongest oxidising agent. It can gain electrons easily. It has highest standard reduction potential.

Zn → Zn²⁺ + 2e^– (oxidation), 

Pb²⁺ → 2e^– +Pb (reduction) 

Zn is oxidized to Zn²⁺ 

Pb²⁺ is reduced to Pb 

Pb²⁺ is the oxidizer

Zn is the reducer.

Electrode potential of Cu²⁺/Cu is higher than H⁺/H₂ and so H⁺ ions cannot oxidize Cu to Cu²⁺ ions and therefore copper does not dissolve in dilute HCl. On the other hand, the electrode potential of NO₃^- ion i.e., NO₃^-/NO electrode is higher than that of copper electrode and hence it can oxidize Cu to Cu²⁺ ion; so copper dissolve in dilute HNO₃. 

Thus copper dissolved in dilute HNO₃ due to oxidation of Cu by NO₃^- ions and not by H⁺ ions.

a. H₂SO₄ :

Oxidation number of H = +1 

Oxidation number of O = -2

H₂SO₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0

∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0

∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0

∴ Oxidation number of S + 2 – 8 = 0

∴ Oxidation number of S in H₂SO₄ = +6

b. HNO₃ :

Oxidation number of H = +1 

Oxidation number of O = -2

HNO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0

∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0

∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0 

∴ Oxidation number of N + 1 – 6 = 0 

∴ Oxidation number of N in HNO₃ = +5

c. H₃PO₃ :

Oxidation number of O = -2 

Oxidation number of H = +1 

H₃PO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0 

∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0 

∴ Oxidation number of P + 3 – 6 = 0 

Oxidation number of P is H₃PO₃ = +3

d. K₂C₂O₄ :

Oxidation number of K = +1 

Oxidation number of O = -2

K₂C₂O₄ is a neutral molecule.

∴ Sum of the oxidation number of all atoms = 0 

∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0 

∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0 

∴ 2 × (Oxidation number of C) + 2 – 8 = 0 

∴ 2 × (Oxidation number of C) = + 6

∴ Oxidation number of C = \(+\frac{6}{2}\)

∴ Oxidation number of C in K₂C₂O₄ = +3

e. H₂S₄O₆ :

Oxidation number of H = +1 

Oxidation number of O = -2

H₂S₄O₆ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0

∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0 

∴ 4 × (Oxidation number of S) + 2 – 12 = 0 

∴ 4 × (Oxidation number of S) = + 10

∴ Oxidation number of S = \(+\frac{10}{4}\)

∴ Oxidation number of S in H₂S₄O₆ = +2.5

f. Cr₂O₇^2- :

Oxidation of O = -2

Cr₂O₇^2- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 2 

∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2 

∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2 

∴ 2 × (Oxidation number of Cr) – 14 = – 2 

∴ 2 × (Oxidation number of Cr) = – 2 + 14 

∴ Oxidation number of Cr = \(+\frac{12}{2}\) 

∴ Oxidation number of Cr in Cr₂O₇^2- = +6

g. NaH₂PO₄ :

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O = -2

NaH₂PO₄ is a neutral molecule

Sum of the oxidation numbers of all atoms = 0 

(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0 (+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0

(Oxidation number of P) + 3 – 8 = 0

Oxidation number of P in NaH₂PO₄ = +5

Option : a. Reduction

Mg → Mg²⁺ + 2e^– (oxidation) 

Cl₂ + 2e^– → 2Cl (reduction) 

Mg is oxidized to Mg 

Cl₂ is reduced to Cl 

Mg is the reducer

Cl₂ is the oxidizer.

Since reduction potential of silver is more than that of hydrogen (E°H⁺/H₂Pt = 0), silver vessel will be suitable to store 1M HCl. On the other hand E° of Al³⁺/Al is less than that of hydrogen (E°H⁺/H₂Pt) so that hydrogen will be liberated is stored in aluminium vessel.

1. Oxidant : 

Oxidant or oxidising agent is an electron acceptor. 

2. Reductant : 

Reductant or reducing agent is an electron donor.

1. Oxidant :

Oxidant or oxidising agent is a substance which increases the oxidation number of an element in a given substance, and itself undergoes decrease in oxidation number of a constituent element in it.

2. Reductant : 

Reductant or reducing agent is a substance that lowers the oxidation number of an element in a given substance, and itself undergoes an increase in the oxidation number of a constituent element in it.

Let oxidation number of Fe = x

\(\therefore\) 3(x) + 4(– 2) = 0

3x – 8 = 0

3x = 8

x = \(+\frac{8}{3}\)

By stoichiometry, Fe₃O₄ = \(\overset{+2-2}{FeO}.\overset{+3}{Fe_2}\overset{-2}{O_3}\)

\(\therefore\) Oxidation number of Fe in Fe₃O₄ is + 2 and +3.

A reagent/substance which itself undergoes oxidation bringing about the reduction of another species is called reductant/reducing agent.

Reducing Agent 

The substance that loses electrons and its valency thereby oxidised to a higher valency state is called a reducing agent.

Combustion of methane : 

CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Combustion is a process in which a substance combines with oxygen.

Batteries generate electricity by redox reactions. 

Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. 

Hence,

New batteries become useless after some days.

Car bumper is made of iron which undergoes rusting over a period of time. 

Hence,

Old car bumper changes colour.

Option : d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Option : b. A₃(BC₄)₂

(D) bromine is both reduced and oxidised

Correct option is: (A) redox

A chemical reaction in which oxidation and reduction processes takes place simultaneously is known as redox reaction

Correct option is: (A) redox

To complete the electric circuit without mixing the two solution of two half cells. It avoids the accumulation of electric charges in two half – cells.

For determining the value of n, the oxidation number of Mn in left and right sides is calculated:

In MnO₄^-, 

x + 4(-2) = -1  (Let oxidation no. of Mn = x)

x - 8 = -1

x = -1 + 8 = +7

In Mn²⁺ 

x = +2

Since, there is decrease in oxidation number of 5. So, there is gain of 5 electrons in this reaction. Hence, the value of n = 5.

No, we cannot use KCI as electrolyte in the salt bridge of the given cell because Cl^– ions will combine with Ag⁺ ions to form white precipitate of AgCI.

Cut apple turns brown when exposed to air because polyphenols are released. 

These polyphenols undergo oxidation in the presence of air and impart brown colour.

Option :  c. – \(\frac{1}{2}\)

Correct option is: (B) -1

The oxidation number of oxygen in peroxide is -1

Correct option is: (B) -1

In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidizing agent.

(i) Among the oxoanions of chlorine, CIO₄^– does not disproportionate because in this oxoanion. Chlorine is present in its highest oxidation state i.e. +7. 

The disproportionation reactions for other there oxoanions of chlorine are as follows:

\(3\overset{+1}{CI}O^-\rightarrow2\overset{-1}{Cl}+\overset{+5}{CIO^-_3}\)

\(6\overset{+3}{CI}O^-_2 \overset{hr}{\rightarrow}4\overset{+5}{CI}O^-_3+2\overset{-1}{Cl^-}\)

\(4\overset{+5}{CI}O^-_3\rightarrow\overset{-1}{Cl^-}+3\overset{+7}{CI}O^-_4\)

(ii) (a) Example of combination reaction

3Mg(s) + N₂(g) → Mg₃N₂(s)

(b) Example of Decomposition reaction:

2NaH(s) \(\overset{\Delta}{\rightarrow}\) 2Na(s) + H₂(g)

(c) Example of metal displacement reaction:

V₂O₅(s) + 5Ca(s) \(\overset{\Delta}{\rightarrow}\) 2V(s) + 5CaO(s)

A disproportionation reaction is an oxidation-reduction process in which the same substance is oxidized and reduced.

e.g., 2Cu⁺ → Cu + Cu²⁺

Balancing of Redox Reaction by half-Reaction Method:

Step-1: First we write the skeletal ionic equation

\(\overset{+6}{Cr_2}\overset{2-}{O_7} (aq)+\overset{+4}{S}\,\overset{2-}{O_3}(aq)\rightarrow \overset{+3}{Cr^{3+}}(aq)+\overset{+6}{S}\,\overset{2-}{O_4}\)

Step-2: The two half reactions are: Oxidation half reaction

\(\overset{+4}{S}\,\overset{2-}{O_3}(aq)\rightarrow \overset{+6}{S}\,\overset{2-}{O_4}(aq)\)

Reduction half reaction

\(\overset{+6}{Cr_2}\overset{2-}{O_7}_{(aq)}\rightarrow \overset{+3}{Cr^{3+}}_{(aq)}\)

Step-3: Balance the Cr atoms in reduction half reaction:

Cr₂O₇^-2_(aq) → 2Cr³⁺ _(aq)

Step-4: To balance O-atoms in oxidation half reaction, we add one water molecule to the left side:

SO₃^2- (aq) + H₂O(I) → SO₄²⁻ (aq)

And to balance O-atoms in reduction-half reaction, we add seven water molecules to the right side:

Cr₂O₇^-2_(aq)→ 2Cr³⁺ _(aq) + 7H₂O(l)

To balance the H-atoms in oxidation half reaction, we add two H⁺ ions to the right side.

SO₃^2- (aq) + H₂O (l) → SO₄²⁻(aq) + 2H⁺(aq)

To, balance the H-atoms in reduction half reaction, we add 14H⁺ ions to be left side.

Cr₂O₇^-2_(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l)

Step-5: Balance the charges of two half reactions,

\(\overset{+4}{S}\overset{2-}{O_3}(aq) +H_2O(l)\rightarrow\overset{+6}{S}\overset{2-}{O_4}(aq) +2H^+(aq) +2e^-\)

\(\overset{+6}{Cr_2}\overset{2-}{O_7}(aq) + 14H^+(aq)+6e^- \) → \(2\overset{+3}{Cr^{3+}}\) (aq) + 7H₂O(I)

Now, to equalize the number of electrons, we multiply the oxidation half reaction by 3, then we get

3SO₃²⁻ (aq) + 3H₂O(I) → 3SO₄²⁻ (aq) + 6H⁺(aq) + 6e⁻

Cr₂O₇²⁻ (aq) + 14⁺ + 6e⁻ + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(I)

Step-6: Add two half reactions to obtain the net reaction after cancelling the electrons and other species on both sides. 

The net equation is:

3SO₃²⁻ (aq) + Cr₂O₇²⁻ (aq) + 8H⁺(aq) → 3SO₄²⁻ (aq) + 2Cr³⁺ (aq) + 4H₂O(I)

A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.

MnO₂ 

Mn + 2x -2 = 0 

Mn + 4 = 0

Mn = +4 

Mn₂O₃ 

2Mn + 2x -2 = 0 

2Mn + -4 = 0 

2 Mn = +4 

Mn = +2 

Mn₂O₇ 

2Mn + -2×7 = 0 

2Mn + -14 = 0 

2Mn = +14 

Mn = +7

Na₂O₂

Let, the oxidation state of O = x

2(+1) + 2(x) = 0

2 + 2x = 0

2x = – 2

x = \(-\frac{2}{2}\) = -1

The reaction which involves the process of oxidation and reduction simultaneously is called redox reaction.

e.g. 2Na(s) + Cl₂(g) → 2NaCl(s)

Cr₂O₇^2-

Let, the oxidation number of Cr = x

2(x) + 7(– 2) = – 2

2x – 14 = – 2

2x = – 2 + 14

x = \(\frac{12}{2}\) = +6

H₂S is oxidized because a more electronegative element, Chlorine is added to hydrogen (or more electropositive element hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.

Removal of oxygen / electronegative element form a substance or addition of hydrogen / electropositive element to a substance.