Apr 18, 2021

# Chemistry Cha 12 Electrochemistry 11th Notes KPK

## Short Questions Electrochemistry11th Notes for kpk

**Q.6 i) What is the oxidation number of N in N2O, NO2, N2O3, HNO3 and NH4+.**

**Answer:**

**NO2**

Oxidation number of O = -2

The oxidation number of N = X

Putting these values in the formula NO2,

X + 2(-2) = 0

X — 4 = 0

X = +4

Hence, the oxidation number of N in NO2 is +4.

**N2O**

Oxidation number of O = -2

Oxidation number of N = X

Putting these values in the formula N2O,

2X + (-2) = 0

2X — 2 = 0

2X = +2

X = +2/2 = +1

Hence, the oxidation state of N in N2O is +1.

**N2O3**

Oxidation number of O = -2

Oxidation number of N = X

Putting these values in the formula N2O3,

2X + 3(-2) = 0

2X — 6 = 0

2X = +6

X = +6/2 = +3

Hence, the oxidation number of N in N2O3 is +3.**HNO3**

Oxidation number of H = +1

Oxidation number of O = -2

Oxidation number of N = X

Putting these values in the formula HNO3,

(+1) + X + 3(-2) = 0

X — 5 = 0

X = +5

Hence, the oxidation state of N in HNO3 is +5.

**NH4+**

Oxidation number of H = +1

Oxidation number of N = X

Putting these values in the formula NH4+,

X + 4(+1) = 1

X — 4 = 1

X = 1–4 = -3

Hence, the oxidation state of N in NH4+ is -3.

**Q.6 ii) What is oxidation state? Give the oxidation number of Mn in KMnO4, MnO2, Mn2O7, MnO4 and K2MnO3.**

**Answer:Oxidation State:**

“The apparent charge (positive or negative) which the atom would have in a molecule or ion is called oxidation state”. Oxidation state only shows the positive or negative character of the atom.

**Calculation of Oxidation Number of Mn:**

Oxidation number of K = +1

Oxidation number of O = -2

Oxidation number of Mn = X

**KMnO4**

Putting the values of oxidation numbers in the formula KMnO4,

(+1) + X + 4(-2) = 0

X — 7 = 0

X = +7

Hence, the oxidation number of Mn in KMnO4 is +7.**MnO2**

Putting the values of oxidation numbers in the formula MnO2,

X + 2(-2) = 0

X — 4 = 0

X = +4

Hence, the oxidation number of Mn in MnO2 is +4.**Mn2O7**

Putting the values of oxidation numbers in the formula Mn2O7,

2X + 7(-2) = 0

2X — 14 = 0

2X = +14

X = +14/2 = +7

Hence, the oxidation number of Mn in Mn2O7 is +7.**MnO4**

Putting the values of oxidation numbers in the formula MnO4,

X + 4(-2) = 0

X — 8 = 0

X = +8

Hence, the oxidation number of Mn in MnO4 is +8.**K2MnO3**

Putting the values of oxidation numbers in the formula K2MnO3,

2(+1) + X + 3(-2) = 0

X — 4 = 0

X = +4

Hence, the oxidation number of Mn in K2MnO3 is +4.

Read more: Chemistry Cha 11 Thermochemistry Class 11th Notes 1st years

**Q.6 iii) In Redox reactions the number of electrons lost is equal to the number of electrons gained. Explain.**

**Answer:**

Redox reactions are those reaction in which both oxidation and reduction takes place. In oxidation, the electrons are removed from a substance and in reduction the electron are added to a substance. As overall redox reaction does not involve electrons it means the number of electrons lost are equal to number of electrons gained.

2Na → 2Na+ + 2e- (oxidation)

Cl2 +2e- → 2Cl- (reduction)

2Na + Cl2 → 2Na+ + 2Cl- (redox reaction)

Two electrons are lost by sodium atoms which are gained by chloride ions in this redox reaction

**Q.6 iv) **Na+ is an oxidizing but Na is a reducing agent

i) The lead storage battery is rechargeable.

ii) A solution of sugar is non-conductor but that of table salt is a good conductor.

iii) A dry cell is not connected to an external source of current.

iv) NaCl is non-conductor in the solid-state but is a good conductor in molten form.

**Answer:**

The oxidizing agent is a substance which oxidizes other substances. As Na+ has a deficiency of electrons due to positive charge on it. It reacts with other substances and oxidizes them by removing electrons. Hence Na+ is an oxidizing agent. On the other hand, the reducing agent is a substance which reduces other substances by adding electrons to them. Being a metal Na readily loses its electrons and reduces other substances by adding electrons to them. Hence Na is a reducing agent.**i.**

The lead storage battery is rechargeable. If the external source is allowed to flow in the opposite direction during recharging all the chemical changes which occurred during discharging are reversed. The PbSO4 and H2O reforms Pb, PbO2 and H2SO4 and the battery is again ready for generating useful electricity.**ii.**

Sugar molecules are not ionized in water, they exist as neutral molecules. So the sugar solution is non-conductor. On the other, table salt (NaCl) is ionized to Na+ and Cl- ions in the solution. The solution then conducts electricity due to the presence of these ions. Hence, a solution of table salt is a good conductor.**iii.**

A dry cell is non-rechargeable which means the changes which occurred during discharging cannot be reversed by applying an external circuit in opposite direction. So a dry cell once discharged is discarded and is never connected to an external circuit to recharge it.**iv.**

In solid-state NaCl is non-conductor. The reason is that there are strong electrostatic forces present between Na+ and Cl- ions in solid-state and the ions are not free to conduct electricity. But in molten state ions become free and the electricity is conducted due to the presence of these free ions.

**Q.6 v) During electrolysis of fused NaCl, sodium metal is collected at the cathode and not at the anode.**

**Answer:**

During electrolysis, the ions always move toward electrodes having opposite charge, i.e. positive ions towards the cathode and negative ions towards the anode. Fused NaCl has Na+ and Cl- ions present in it. Positive sodium ions Na+ are attracted by the cathode and get discharged on it forming sodium metal. That is why sodium is always collected at the cathode.

Na+ → Na + e-

**Q.6 vi) SHE acts as anode with Cu electrode but acts as cathode when connected to Zn electrode.**

**Answer:**

The element which has the higher value of reduction potential acts as a cathode and the other with lower value always acts as anode. SHE has standard reduction potential equal to zero. Cu has greater potential to get reduced than SHE and has a positive value of reduction potential. That is why it always acts as cathode when connected with SHE and SHE itself acts as anode. When SHE is connected with Zn, it acts as cathode because reduction potential of Zn is negative. Due to negative reduction potential Zn tends to get oxidized acting as anode. So the SHE acts as cathode when connected with Zn.

**Q.6 vii) In electroplating, the article to be plated is made cathode.**

**Answer:**

In the process of electroplating the article to be plated is made cathode and the pure metal is made anode. The article to be plated is made cathode because metallic ions are positive and thus get deposited on the cathode. Positive metal ions are attracted by cathode (-ve charge), get reduced on cathode and deposit forming a smooth layer of metal.

**Q.6 viii) Write the half cell reactions for the cell, Al /Al3+(1M) // Cu2+(1M) / Cu.**

**Answer:**

Cu has higher reduction potential than Al. Hence, Cu will act as cathode and Al as an anode. Half cell reactions will be the following,

At Cathode Cu+2 + 2e- → Cu (reduction)

At Anode Al → Al+3 + 3e- (oxidation)

## Chemistry 11th Notes for KPK Long Questions

Q.2) Balance the following Redox Equations by Oxidation Number Method

**i) KMnO**4** + H**2**S + H**2**SO**4** → KHSO**4** + MnSO**4** + S + H**2**O****ii) Fe + V**2**O**3** → Fe**2**O**3** + VO****iii) MnO**2** + HCl → MnCl**2** + CI**2** + H**2**O****iv) KMnO**4** + KNO**2** + H**2**SO**4** → MnSO**4** + KNO**3** + K**2**SO**4** + H**2**O****v) Mg + HCl → MgCl**2** + H**2**vi) Cu + H**2**SO**4 **→ CuSO**4** + SO**2** + H**2**O**

**Answer:**

**i. KMnO4 + H2S + H2SO4 → KHSO4 + MnSO4 + S + H2O****Solution:**

(i) KMnO4 + H2S + H2SO4 → KHSO4 + MnSO4 + S + H2O

(ii) Oxidation number of Mn in KMnO4 is +7, the oxidation number of Mn in MnSO4 is +2. The oxidation number of Mn decreases from +7 to +2 so it is reduced and acts as an oxidizing agent. The oxidation number of S increases from -2 in H2S to 0 in S, so it is oxidized and acts as a reducing agent.

(iii) Writing the oxidation number over the symbols of the elements oxidized and reduced.

(iv) Indicating the change in oxidation number by means of arrows.

Hence the substances reduced and oxidized are written as follows.

Multiplying equation (a) by 2 and equation (b) by 3 so in order to balance the number of electrons lost and gained.

2KMnO4 + 5H2S + H2SO4 → KHSO4 + 2MnSO4 + 5S + H2O

(v) Balance K atoms by multiplying KHSO4 by 2.

2KMnO4 + 5H2S + H2SO4 → 2KHSO4 + 2MnSO4 + 5S + H2O

(vi) Balance SO42- ions by multiplying H2SO4 by 4.

2KMnO4 + 5H2S + 4H2SO4 → 2KHSO4 + 2MnSO4 + 5S + H2O

(vii) O atoms are balanced by multiplying H2O by 8. H atoms also get balanced this way. The final balanced equation is as follows.

2KMnO4 + 5H2S + 4H2SO4 → 2KHSO4 + 2MnSO4 + 5S + 8H2O

**ii. Fe + V2O3 → Fe2O3 + VO**Solution:

(i) Fe + V2O3 → Fe2O3 + VO

(ii) Oxidation state of Fe is zero and the oxidation state of Fe in Fe2O3 is +3. The oxidation number of Fe increases from 0 to +3 so it is oxidized and acts as reducing agent. The oxidation number of V in V2O3 is +3 and in VO it is +2. So V is reduced from +3 to +2.

(iii) Writing the oxidation numbers above elements which are oxidized or reduced.

(iv) Indicating the change in oxidation numbers by means of arrows.

Balancing the atoms which are reduced or oxidized,

In order to balance the number of electrons gained and lost, multiply equation (b) by 3.

So the equation takes the form,

2Fe + 3V2O3 ⟶ Fe2O3 + 6VO

(v) Oxygen atoms are already balanced. So the final balance equation is the following.

2Fe + 3V2O3 ⟶ Fe2O3 + 6VO

**iii. MnO2 + HCl → MnCl2 + CI2 + H2O****Solution:**

(i) MnO2 + HCl → MnCl2 + CI2 + H2O

(ii) Oxidation state of Mn is +4 in MnO2 and +2 in MnCl2. So the oxidation state of Mn in changed from +4 to +2 and it is reduced hence acting as an oxidizing agent. Cl is oxidized from -1 in HCl to 0 in Cl2.

(iii) Writing the oxidation numbers above elements which are reduced or oxidized. HCl is written twice on left side because in MnCl2, the oxidation state of Cl remains the same while it is changed in Cl2.

Electrons gained and lost are also balanced this way. The following equation is obtained as a result.

HCl + MnO2 + 2HCl → MnCl2 + CI2 + H2O

(vi) Balancing Cl atoms which are not oxidized by multiplying HCl (not oxidized) by 2.

2HCl + MnO2 + 2HCl → MnCl2 + CI2 + H2O

(vii) Balancing H atoms by multiplying H2O by 2.

2HCl + MnO2 + 2HCl → MnCl2 + CI2 + 2H2O

Combining HCl molecules, the resulting balanced equation is obtained as following.

MnO2 + 4HCl → MnCl2 + CI2 + 2H2O

**iv. KMnO4 + KNO2 + H2SO4 → MnSO4 + KNO3 + K2SO4 + H2OSolution:**

(i) KMnO4 + KNO2 + H2SO4 → MnSO4 + KNO3 + K2SO4 + H2O

(ii) Oxidation number of Mn in KMnO4 is +7 while in MnSO4 it is +2. So it is oxidized from +7 to +2 and hence acts as reducing agent. Nitrogen is reduced from +3 in KNO2 to +5 in KNO3.

(iii) Writing the oxidation states of the elements which are oxidized or reduced.

The resulting equation takes the following form,

2KMnO4 + 5KNO2 + H2SO4 → 2MnSO4 + 5KNO3 + K2SO4 + H2O

(vi) Balancing SO4–2 ions by multiplying H2SO4 by 3.

2KMnO4 + 5KNO2 + 3H2SO4 → 2MnSO4 + 5KNO3 + K2SO4 + H2O

(vii) H atoms are balance by multiplying H2O by 3. Final balanced equation is obtained as follows.

2KMnO4 + 5KNO2 + 3H2SO4 → 2MnSO4 + 5KNO3 + K2SO4 + 3H2O

**v. Mg + HCl → MgCl2 + H2Solution:**

(i) Mg + HCl → MgCl2 + H2

(ii) Mg is oxidized as its oxidation state is changed from 0 in Mg to +2 MgCl2. Hydrogen is reduced from +1 in HCl to 0 in H2.

(iii) Writing the oxidation states of the elements which are oxidized or reduced.

Electrons gained and lost are also balanced thus. The resulting equation takes the following form.

Mg + 2HCl → MgCl2 + H2

All the other atoms are balanced, so this is the balanced chemical equation.

**vi. Cu + H2SO4 → CuSO4 + SO2 + H2OSolution:**

(i) Cu + H2SO4 → CuSO4 + SO2 + H2O

(ii) Cu is oxidized from oxidation state 0 in Cu to +2 in CuSO4. Sulphur is reduced from +6 in H2SO4 to +4 in SO2.

(iii) Writing the oxidation states of the elements which are oxidized or reduced.

(iv) The elements which are undergone oxidation or reduction are shown with the means of arrows. Sulphur is reduced in SO2 while it remains unchanged in CuSO4, so H2SO4 on left side is written twice.

Oxidation and reduction are taking place as follows.

Electrons gained and lost are already balanced. The equation takes the following form.

H2SO4 + Cu + H2SO4 → CuSO4 + SO2 + H2O

Cu + 2H2SO4 → CuSO4 + SO2 + H2O

(v) Balancing hydrogen atoms by multiplying H2O by 2. The final balanced equation is as follows.

Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

**Q.3) Balance the following Redox Equations by the Half Reaction Method**

i) Sn2+ + Fe3+ → Sn4+ + Fe2+

ii) Zn + Cr2O72- + H+ → Zn2+ + Cr3++ H2O

iii) H2O2 + MnO4- + H+→ Mn2++ O2+H2O

iv) IO3- + AsO33- → l- + AsO43- (basic medium)

v) Sn2+ + I2 → Sn4++ l-

vi) l- + OCl- → I2 + Cl- + H2O (basic medium)

vii) CN- + MnO4- →CNO- + MnO2 (basic medium)

**Answer:**

**i. Sn2+ + Fe3+ → Sn4+ + Fe2+****Solution:**

1. Split the equation into two half-reactions, one for oxidation and the other for reduction.

(i) Sn2+ ⟶ Sn4+ (Oxidation)

(ii) Fe3+ ⟶ Fe2+ (Reduction)

2. Atoms on both sides of both of the half-reactions are already balanced.

3. Balance charge on each side of both half-reactions by adding electrons to each side.

(i) Sn2+ ⟶ Sn4+ + 2e-

(ii) Fe3+ + e- ⟶ Fe2+

4. Multiply equation (ii) by 2 in order to balance the electrons on the two sides of the half-reactions. Add two reactions after cancelling the common species in the two reactions.

(i) Sn2+ ⟶ Sn4+ + 2e-

(ii) 2Fe3+ + 2e- ⟶ 2Fe2+

_______________________________________

Sn2+ + 2Fe3+ ⟶ Sn4+ + 2Fe2 (Balanced equation)

**ii. Zn + Cr2O72- + H+ → Zn2+ + Cr3++ H2O****Solution:**

1. Split the equation into two half-reactions, one for oxidation and the other for reduction.

(i) Zn ⟶ Zn2+ (Oxidation)

(ii) Cr2O72- ⟶ Cr3+ (Reduction)

2. Balance atoms on both sides of the half-reactions by using H2O and H+ species.

(i) Zn ⟶ Zn2+

(ii) Cr2O72- + 14H+ ⟶ 2Cr3+ + 7H2O

3. Balance charge on each side of both half-reactions by adding electrons to each side.

(i) Zn ⟶ Zn2+ + 2e-

(ii) Cr2O72- + 14H+ + 6e- ⟶ 2Cr3+ + 7H2O

4. Multiply equation (i) by 3 to balance the electrons in two half-reactions. Then adding (i) and (ii) after cancelling the common species in the two reactions.

(i) 3Zn ⟶ 3Zn2+ + 6e-

(ii) Cr2O72- + 14H+ + 6e- ⟶ 2Cr3+ + 7H2O

_______________________________________

3Zn + Cr2O72- + 14H+ ⟶ 3Zn2+ + 2Cr3+ + 7H2O (Balanced equation)

**iii. H2O2 + MnO4- + H+→ Mn2++ O2+ H2OSolution:**

1. Split the equation into two half reactions, one for oxidation and the other for reduction.

(i) H2O2 ⟶ O2 (Oxidation)

(ii) MnO4- ⟶ Mn2+ (Reduction)

2. Balance atoms on both sides of the half reactions by using H2O and H+ species.

(i) H2O2 ⟶ O2 + 2H+

(ii) MnO4- + 8H+ ⟶ Mn2+ + 4H+O

3.