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) KMnO₄ + H₂S + H₂SO₄ → KHSO₄ + MnSO₄ + S + H₂O
ii) Fe + V₂O₃ → Fe₂O₃ + VO
iii) MnO₂ + HCl →  MnCl₂ + CI₂ + H₂O
iv) KMnO₄ + KNO₂ + H₂SO₄ →  MnSO₄ + KNO₃ + K₂SO₄ + H₂O
v) Mg + HCl → MgCl₂ + H₂
vi) Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O

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 SO₄^2- 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 + H₂O
Solution:
(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 + H₂
Solution:
(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 + H₂O
Solution:
(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 + Cr₂O₇^2- + H⁺ → Zn²⁺ + Cr³⁺+ H₂O
iii) H₂O₂ + MnO₄^– + H⁺→ Mn²⁺+ O₂+H₂O
iv) IO₃^– + AsO₃^3- → l^– + AsO₄^3-                                 (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 + Cr₂O₇^2- + 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) Cr₂O₇^2- ⟶ 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) Cr₂O₇^2- + 14H⁺ + 6e^– ⟶ 2Cr³⁺ + 7H₂O
_______________________________________
3Zn + Cr₂O₇^2- + 14H⁺ ⟶ 3Zn²⁺ + 2Cr³⁺ + 7H₂O (Balanced equation)

iii. H₂O₂ + MnO₄^– + H⁺→ Mn²⁺+ O₂+ H₂O
Solution:
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. Balance charge on each side of both half reactions by adding electrons to each side.
(i) H₂O₂ ⟶ O₂ + 2H⁺ + 2e^–
(ii) MnO₄^– + 8H⁺ + 5e^‑⟶ Mn²⁺ + 4H₂O
4. Multiply equation (i) by 5 and equation (ii) by 2 in order to balance the electrons in two half reactions. Then adding (i) and (ii) after cancelling the common species in the two reactions.
(i) 5H₂O₂ ⟶ 5O₂ + 10H⁺ + 10e^–
(ii) 2MnO₄^– + 16H⁺ + 10e^‑ ⟶ 2Mn²⁺ + 8H₂O
_________________________________________
5H₂O₂ + 2MnO₄^– + 6H⁺ ⟶ 5O₂ + 2Mn²⁺ + 8H₂O (Balanced equation)

iv. IO₃^– + AsO₃^3- → l^– + AsO₄^3-                                       (basic medium)
Solution:
1. Split the equation into two half reactions, one for oxidation and the other for reduction.
(i) AsO₃^3- ⟶ AsO₄^3- (Oxidation)
(ii) IO₃^– ⟶ I^– (Reduction)
2. As the reaction is taking place in basic medium. Balance atoms on both sides of the half reactions by using H₂O and OH^– species.
(i) AsO₃^3- + 2OH^– ⟶ AsO₄^3- + H₂O
(ii) IO₃^– + 3H₂O ⟶ I^– + 6OH^–
3. Balance charge on each side of both half reactions by adding electrons to each side.
(i) AsO₃^3- + 2OH^– ⟶ AsO₄^3- + H₂O + 2e^–
(ii) IO₃^– + 3H₂O + 6e^– ⟶ I^– + 6OH^–
4. Multiply equation (i) by 3 in order to balance the electrons in both sides of the half reactions. Then adding equation (i) and (ii) after cancelling the common species.
(i) 3AsO₃^3- + 6OH^– ⟶ 3AsO₄^3- + 3 H₂O + 6e^–
(ii) IO₃^– + 3H₂O + 6e^– ⟶ I^– + 6OH^–
_______________________________________
3AsO₃^3- + IO₃^– ⟶ 3AsO₄^3- + I^– (Balanced equation)

v. Sn²⁺ + I₂ → Sn⁴⁺+l^–
Solution:
1. Split the equation into two half reactions, one for oxidation and the other for reduction.
(i) Sn²⁺ ⟶ Sn⁴⁺ (Oxidation)
(ii) I₂ ⟶ 2I^– (Reduction)
2. Balance atoms on both sides of the half reactions.
(i) Sn²⁺ ⟶ Sn⁴⁺
(ii) I₂ ⟶ 2I^–
3. Balance charge on each side of both half reactions by adding electrons to each side.
(i) Sn²⁺ ⟶ Sn⁴⁺ + 2e^–
(ii) I₂ + 2e^– ⟶ 2I^–
4. Electrons on both sides of the two half reactions are already balanced. Adding equation (i) and (ii) to obtain the final balanced equation,
(i) Sn²⁺ ⟶ Sn⁴⁺ + 2e^–
(ii) I₂ + 2e^– ⟶ 2I^–
______________________
Sn²⁺ + I₂ ⟶ Sn⁴⁺ + 2I^–  (Balanced equation)

vi. I^– + OCl^–  → I₂ + Cl^– + H₂O                                          (basic medium)
Solution:
1. Split the equation into two half reactions, one for oxidation and the other for reduction.
(i) I^– ⟶ I₂ (Oxidation)
(ii) OCl^– ⟶ Cl^– (Reduction)
2. As the reaction is taking place in basic medium, so we use OH^– and H₂O species in order to balance the atoms on both sides of the two half reactions.
(i) 2I^– ⟶ I₂
(ii) OCl^– + H₂O ⟶ Cl^– + 2OH^–
3. Balance charge on each side of both half reactions by adding electrons to each side.
(i) 2I^– ⟶ I₂ + 2e^–
(ii) OCl^– + H₂O + 2e^‑ ⟶ Cl^– + 2OH^–
4. Electrons on each side of the two half reactions are already balanced. Adding equation (i) and (ii) to obtain the final balanced equation,
(i) 2I^– ⟶ I₂ + 2e_–
(ii) OCl^– + H₂O + 2e^‑ ⟶ Cl^– + 2OH^–
___________________________________
2I^– + OCl^– + H₂O ⟶ I₂ + Cl^– + 2OH^– (Balanced equation)

vii. CN^– + MnO₄^–  → CNO^– + MnO₂                                 (basic medium)
Solution:
1. Split the equation into two half-reactions, one for oxidation and the other for reduction.
(i) CN^‑ ⟶ CNO^– (Oxidation)
(ii) MnO₄^– ⟶ MnO₂ (Reduction)
2. As the reaction is taking place in basic medium, so we use OH^– and H₂O species in order to balance the atoms on both sides of the two half reactions.
(i) CN^‑ + 2OH^– ⟶ CNO^– + H₂O
(ii) MnO₄^– + 2H₂O ⟶ MnO₂ + 4OH^–
3. Balance charge on each side of both half reactions by adding electrons to each side.
(i) CN^‑ + 2OH^– ⟶ CNO^– + H₂O + 2e^–
(ii) MnO₄^– + 2H₂O + 3e^– ⟶ MnO₂ + 4OH^–
4. Multiply equation (i) by 3 and equation (ii) by 2 in order to balance electrons on each side of the two half reactions. Then adding equation (i) and (ii) after cancelling the common species to get the final balanced equation.
(i) 3CN^‑ + 6OH^– ⟶ 3CNO^– + 3H₂O + 6e-
(ii) 2MnO₄^– + 4H₂O + 6e^– ⟶ 2MnO₂ + 8OH^–
______________________________________
3CN^‑ + 2MnO₄^– + H₂O ⟶ 3CNO^– + 2MnO₂ + 2OH^–

Q.4) Given the following cells at 25°C.

a) Write the cell reactions              
b) Calculate the cell voltage
i. 1) Mg²⁺/ Mg                   E°Red= -2.38 V
    2) Cu²⁺/ Cu                    E°Red = +0.34 V
ii.1) 2H⁺/ H₂                       E°Red = 0.00 V
    2) Ag⁺/ Ag                       E°Red = +0.80 V

Answer:

i)
As the reduction potential of Cu is greater than that of Mg, so Cu will act as cathode and reduction will take place on it while Mg will act as anode and oxidation will take place on it. The corresponding oxidation-reduction reaction will be the following.
At Cathode       Cu⁺² + 2e- → Cu    (reduction)
At Anode          Mg → Mg⁺² + 2e^– (oxidation)
As Mg is force to undergo oxidation, hence its oxidation potential will be +2.38 V. The cell voltage is calculated as follows,
E^o_Cell = E^o_Red + E^o_Oxi = 0.34 V + 2.38 V = 2.72 V
ii)
As the reduction potential of Ag is greater than that of hydrogen, so Ag will act as cathode and reduction will take place on it while hydrogen will act as anode and oxidation will take place on it. The corresponding oxidation-reduction reaction will be the following.
At Cathode       2(Ag⁺ + e^– → Ag)     (reduction)
At Anode           H₂ → 2H⁺ + 2e^–       (oxidation)
As hydrogen is force to undergo oxidation, hence 0 V will be its oxidation potential. The cell voltage is calculated as follows,
E^o_Cell = E^o_Red + E^o_Oxi = 0.80 V + 0 V = 0.80 V

Q.5) What is the net cell reaction and the cell voltage of the following half reactions

i) Fe³⁺ + 3e^– → Fe                     E°_Red = +0.77 V
ii) I₂+ 2e^–  → 2l-                          E°_Red = +0.24 V

Answer:
As Fe⁺³ has a higher reduction potential than iodine, so it will undergo reduction and I₂ will get oxidized.
2(Fe³⁺ + 3e^– → Fe)                         E°_Red = +0.77 V
3(2l^– → I₂+ 2e^–)                             E°_Oxi = -0.24 V
——————————————————
2Fe⁺³ + 6I^– → 2Fe + 3I₂               E^o_Cell = 0.53 V

Q.7 i) a. Define the term electrolysis. State and explain Faraday’s laws of electrolysis.

Answer:
Electrolysis:
“The chemical decomposition of compound produced by passing electric current through its molten or aqueous solution in electrolytic cell is called electrolysis”. Oxidation and reduction reactions take place on respective electrodes during electrolysis process.
Faraday’s Laws of Electrolysis:
Michael Faraday (1813) gave a relationship between the quantity of electricity passed and the amount of substance deposited at the electrode.
First Law:
The amount of any substance (W) deposited or liberated at an electrode is directly proportional to the quantity of electricity (Q) passed.
W α Q
But the quantity of electricity (Q) is the product of current I and time t. so we can write
W α It
W = ZIt
Where Z is proportionality constant.
Second Law:
If the same amount of electricity is passed through different electrolytes, the amounts of different substances deposited are in the ratio of their chemical equivalents.
W α chemical equivalent (e)    (2nd law)
W α It                                       (1st law)
Combining the two laws,
W α Ite
Introducing the proportionality constant 1/F, we get.
W=Ite/F
“F” is called Faraday’s constant which has a value of 96500C.

Q.7 i) b. A certain amount of current is passed through CuSO₄ solution for fifty minutes. The net amount of Cu deposited was found to be 25g. Calculate the current when Z = 1 and At. Mass of Cu is 63.5g.

Answer:
Time = t = 50 minutes = 50 x 60 = 300 s
Amount of Cu deposited = W = 25 g
Z = 1
Gram equivalent of Cu = e = 63.5 g
F = 96500 C
Using Faradays law,
W=Ite/F
Rearranging for I and putting the values,

Q.7 ii) What is an electrode potential? How it can be measured? Explain SHE in detail.

Answer:
Electrode Potential:
“The potential set up when an electrode is in contact with one molar solution of its own ions at 298 K is known as standard electrode potential of the element”. It is represented as E^o.

Standard Hydrogen Electrode (SHE)
A standard hydrogen electrode consists of a platinum foil, coated with a layer of finely divided platinum electrolytically for the purpose to increase its surface area and hence to increase the rate of reaction. The electrode is suspended in 1M solution at 25^oC.
Pure hydrogen is bubbled over the platinum electrode at one atmospheric pressure. The standard electrode potential of any electrode is obtained by combining the electrode with SHE. Since Eo of hydrogen electrode is zero in either case i.e. reduction as well as oxidation so the measured potential is the electrode potential of the other electrode.

Measurement of Electrode Potential
In any measurement of electrode potential the concerned electrode is joined electrically with the standard hydrogen electrode (SHE) and a galvanic cell is established. The two solutions are separated by a porous partition or a salt bridge containing a concentrated solution of potassium chloride. The salt bridge is used to provide a highly conducting path between the two electrolytic solutions. The potential difference is measured by a voltmeter which gives the potential of the electrode as the potential of SHE is zero. An oxidation or reduction may take place at SHE depending upon the nature of the electrode which is coupled with it.

To measure the electrode potential of zinc, a galvanic cell is established between zinc electrode dipped in 1M solution of its ions and standard hydrogen electrode at 25oC. Under the standard conditions the voltmeter reads 0.76 volts and the deflections are in such a direction as to indicate that zinc has a greater tendency to give off electrons than hydrogen. In other words, the half reaction Zn(s) ⟶ Zn²⁺(aq) + 2e^– has greater tendency to occur than H₂(g) ⟶ 2H⁺(aq) + 2e^– by 0.76 volts. The standard potential of zinc is, therefore, 0.76 volts. It is called oxidation potential of Zn and is given the positive sign. The reduction potential of Zn electrode is -0.76 volt. The electrode reactions will be shown as follows.
At anode   Zn(s) ⟶ Zn²⁺(aq) + 2e^– (oxidation)
At cathode 2H+(aq) + 2e^– ⟶ H₂(g) (reduction)

Q.7 iii) What is the difference between the following? Give examples.]

i) Reversible and irreversible cells
ii) Electrolytic and voltaic cell
iii) Electrolytic and electrical conduction

Answer:
i) Difference between Reversible & Irreversible Cells:

 
ii) Difference between Electrolytic & Voltaic Cell:

 
iii) Difference between Electrolytic & Electric conduction:

Q.7 iv) a. Describe the rules used for assigning the oxidation number to an element in a compound.

Answer:
Rules for assigning the oxidation number:

1. The oxidation number of a free element is zero, e.g. oxidation number of H₂, O₂ and Mn is zero.
2. The oxidation number of hydrogen in its compounds is +1 but in metal hydrides it is -1, e.g. NaH and MgH₂.
3. The oxidation number of oxygen in the compounds is -2 but in peroxides it is -1 and in OF₂ is +2.
4. The oxidation number of the elements of group I, II and III in the compounds is +1, +2 and +3 respectively.
5. The oxidation number of halogens of group VII in the binary compounds is -1.
6. The algebraic sum of oxidation numbers all the atoms in a molecule is zero.
7. The algebraic sum of the oxidation numbers of all the atoms in an ion is equal to the charge on the ion.
8. When an atom is oxidized its oxidation number increases and when an atom is reduced, its oxidation number is decreased.

Q.7 iv) b. Calculate the oxidation number of the central atom in the following,

i) Na₂CO₃        
ii) Na₃PO₄          
iii) K₂MnO₃            
iv) HNO₃            
v) KClO₃

Answer:
i) Oxidation Number of C in Na₂CO₃
Oxidation number of Na = +1
Oxidation number of O = -2
Oxidation number of C = X
Putting the values of oxidation number in formula Na₂CO₃,
2(+1) + X + 3(-2) = 0
X – 4 = 0
X = +4
Hence the oxidation number of central atom C in Na₂CO₃ is +4.

ii) Oxidation Number of P in Na₃PO₄
Oxidation number of Na = +1
Oxidation number of O = -2
Oxidation number of P = X
Putting the values of oxidation number in formula Na₃PO₄,
3(+1) + X + 4(-2) = 0
X – 5 = 0
X = +5
Hence the oxidation number of central atom P in Na₃PO₄ is +5.

iii) Oxidation Number of Mn in K₂MnO₃
Oxidation number of K = +1Oxidation number of O = -2
Oxidation number of Mn = X
Putting the values of oxidation number in formula K₂MnO₃,
2(+1) + X + 3(-2) = 0
X – 4 = 0
X = +4
Hence the oxidation number of central atom Mn in K₂MnO₃ is +4.

iv) Oxidation Number of N in HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
Oxidation number of N = X
Putting the values of oxidation number in formula HNO₃,
(+1) + X + 3(-2) = 0
X – 5 = 0
X = +5
Hence the oxidation number of central atom N in HNO₃ is +5.

v) Oxidation Number of Cl in KClO₃
Oxidation number of K = +1
Oxidation number of O = -2
Oxidation number of Cl = X
Putting the values of oxidation number in formula KClO₃,
(+1) + X + 3(-2) = 0
X – 5 = 0
X = +5
Hence the oxidation number of central atom Cl in KClO₃ is +5.

Q.7 v) a. What is an electrolytic cell? Explain Daniell cell in detail.

Answer:
Electrolytic Cell:
“It is a device in which a non-spontaneous chemical reaction is carried out by passing electric current”.
Structure and Working
The electrolytic cell consists of a vessel containing electrolyte in which two metallic plates acting as electrodes, are suspended. The electric current enters and leaves the cell through the electrode. The electrodes connected to the negative terminal of the battery is called cathode while the other connected to the positive terminal is anode.

When an electric current is passed through electrolytic solution, its ions move towards the oppositely charged electrodes. The anions liberate electrons at anode and are said to be oxidized. These electrons pass through outer circuit to the cathode. The cations which surround the cathode, consume those electrons and get deposited at the electrode. Hence the number of electrons lost is always equal to the number of electrons gained.
Daniell Cell
Daniell cell is an example of voltaic cell in which a spontaneous oxidation reduction reaction is carried out and electrical current is produced.
Structure and Working
This cell has a Zn electrode dipped into 1M ZnSO₄ solution and a Cu electrode immersed in a 1M solution of Cu⁺² ions. These half cells are externally connected through a metallic wire while internally they are connected by a salt bridge. The salt bridge contains an aqueous KCl solution in a gel.

Zn tends to lose electrons more readily than Cu giving its electrons to the electrode. Electrons flow from the Zn electrode to the Cu electrode through external circuit making it negatively charged. The Cu²⁺ surrounding the electrons and get deposited as neutral metal atom. Reduction takes place at copper cathode whereas oxidation at Zn electrode. The following half reactions occur at the electrodes.
At Anode    Zn ⟶ Zn²⁺ + 2e^–
At Cathode Cu²⁺ + 2e^– ⟶ Cu
_________________________________________
Net Reaction Zn + Cu²⁺ ⟶ Zn²⁺ + Cu
The cell is represented as follows, Zn(s)/Zn²⁺(aq)(1M)//Cu²⁺(aq)(1M)/Cu(s)

Q.7 v) b. Calculate the electrical energy obtained from a Daniell cell.

Answer:
Electrical Energy from Daniell Cell
Electrical energy (cell voltage) of daniell cell can be calculated by considering the reduction and oxidation reactions occurring at cathode and anode respectively and their corresponding electrode potential. Oxidation takes place at anode and its oxidation potential is +0.76V. Reduction takes place at cathode and its reduction potential is +0.34V. Net voltage of the cell is the sum of these two electrode potentials.
At Anode Zn ⟶ Zn²⁺ + 2e^– E^o_Ox = +0.76V
At Cathode Cu²⁺ + 2e^– ⟶ Cu E^o_Red = +0.34V
_________________________________________
Net Reaction Zn + Cu²⁺ ⟶ Zn²⁺ + Cu E^o_Cell = 1.10 V
So the net voltage of the cell is 1.10 V.

Q.7 vi) a. Explain the terms oxidation and reduction with examples.

Answer:
Oxidation:
“Addition of oxygen, removal of hydrogen or the removal of electrons in any substance is known as oxidation”. According to classical concept, the addition of oxygen and the removal of hydrogen is oxidation. For example carbon is oxidized to CO₂ with the addition of O₂ and ammonia is oxidized to N₂ by the removal of H₂.
C + O₂ ⟶ CO₂
2NH₃ ⟶ N₂ + 3H₂
According to modern concept, the removal of electron from any specie is known as oxidation. For example, sodium (Na) is oxidized to Na⁺ by the removal of electron.
Na ⟶ Na⁺ + e^–
Reduction:
Removal of oxygen, addition of hydrogen or addition of electrons in any substance is known as reduction. According to classical concept, the removal of oxygen and the addition of hydrogen is called reduction. For example, CO₂ is reduced to carbon by the removal of O₂ from it and Cl₂ is reduced to HCl by the addition of hydrogen in it.
CO₂ ⟶ C + O₂
H₂ + Cl₂ ⟶ 2HCl
According to modern concept, the addition of electron in any specie is known as reduction. For example Cl₂ is reduced to Cl^– ions by the addition of electrons.
Cl₂ + 2e^– ⟶ 2Cl^–

Q.7 vi) b.  What are the half cell reactions? Give the half cell reactions of Nelson’s cell for the production of NaOH.

Half-Cell Reaction:
A half cell reaction is a reaction which occurs in a half cell. It is either an oxidation reaction in which electrons are lost or a reduction reaction where electronic is gained. The reactions occur in an electrochemical cell in which the electrons are lost at the anode through oxidation and consumed at the cathode where the reduction occurs.
Nelson’s Cell for NaOH production
NaOH is produced commercially by the electrolysis of aqueous NaCl in Nelson’s cell. Nelson’s cell is an oblong steel tank containing a concentrated aqueous solution of NaCl. The graphite anode is suspended in the solution. The cathode is made of a sheet of perforated steel.

When connected to the battery, the half-cell reactions taking place at electrodes are as follows,
At Anode     2Cl^– ⟶ Cl₂ + 2e^– (oxidation)
At Cathode 2H₂O + 2e^– ⟶ H₂ + 2OH^– (reduction)
__________________________________________
Overall reaction 2Cl^– + 2H₂O ⟶ 2Na⁺ + H₂ + Cl₂ + 2OH^–
Cl₂ is releaseed at anode and H₂ at cathode. NaOH is collected at the bottom of the cell.

Q.7 vii) What are electrochemical cells? Explain it in detail.

Answer:
ELECTROCHEMICAL CELL
“A device in which inter-conversion of electrical and chemical energies takes place is called an electrochemical cell”. Electrochemical cells are ot two types.
Electrolytic Cell: In electrolytic cells, the electrical energy from an external source is used to bring about a chemical change. Electrolytic purification of metals and charging of batteries are its examples.
Voltaic or Galvanic Cell: It is a device in which chemical energy is converted into electrical energy. Dry cells and lead storage battery are its examples.
Electrolytic Cell:
“It is a device in which a non-spontaneous chemical reaction is carried out by passing electric current”.

Structure and Working
The electrolytic cell consists of a vessel containing electrolyte in which two metallic plates acting as electrodes, are suspended. The electric current enters and leaves the cell through the electrode. The electrodes connected to the negative terminal of the battery is called cathode while the other connected to the positive terminal is the anode.

When an electric current is passed through electrolytic solution, its ions move towards the oppositely charged electrodes. The anions liberate electrons at anode and are said to be oxidized. These electrons pass through outer circuit to the cathode. The cations which surround the cathode, consume those electrons and get deposited at the electrode. Hence the number of electrons lost is always equal to the number of electrons gained.

Voltaic Cell:
In the voltaic cell a spontaneous oxidation reduction reaction is carried out and electrical energy is produced. It consists of two half-cells externally connected with a metallic wire acting as a conductor. At each half cell one half of the total cell reaction takes place. At one electrode electrons enter resulting in reduction reaction while at the other they leave the solution and oxidation takes place. A typical example of the voltaic cell is that of Daniell Cell.

Daniell Cell:
Daniell cell is an example of voltaic cell in which a spontaneous oxidation reduction reaction is carried out and electrical current is produced.

Structure and Working
This cell has a Zn electrode dipped into 1M ZnSO₄ solution and a Cu electrode immersed in a 1M solution of Cu⁺² ions. These half cells are externally connected through a metallic wire while internally they are connected by a salt bridge. The salt bridge contains an aqueous KCl solution in a gel.

Zn tends to lose electrons more readily than Cu giving its electrons to the electrode. Electrons flow from the Zn electrode to the Cu electrode through external circuit making it negatively charged. The Cu²⁺ surrounding the electrons and get deposited as neutral metal atom. Reduction takes place at copper cathode whereas oxidation at Zn electrode. The following half reactions occur at the electrodes.
At Anode Zn ⟶ Zn²⁺ + 2e^–
At Cathode Cu²⁺ + 2e^– ⟶ Cu
_________________________________________
Net Reaction Zn + Cu²⁺ ⟶ Zn²⁺ + Cu
The cell is represented as follows, Zn(s)/Zn²⁺(aq)(1M)//Cu²⁺(aq)(1M)/Cu(s)

Q.7 viii)

Answer:

Write down the types of batteries. Explain fuel cells?

Types of Batteries:
Batteries have four major types which are following,

Primary Batteries: These batteries are not reversible and once discharged are discarded. For example dry cell.
Secondary Batteries: These are reversible and can be charged. For example, Lead storage battery.
Solar Batteries: These are photochemical cells and generate energy.

Fuel Cells:
“An electrochemical device for continuously converting chemicals (a fuel and an oxidant) into direct current (D.C.) is called fuel cell”.
Construction and Working:
It consists of two electronic electrodes separated by an ionic electrolyte with the provision for the continuous movement of fuel, oxidant and reaction products into and out of the cell. The fuel can be gas, liquid, solid or solution. The electrodes may be solid or porous and may contain a catalyst. Fuel cells differ from common batteries that in this electricity is produced from chemical fuels fed to them as needed so that their operating life is unlimited.
Fuel cell-based on hydrogen and oxygen has a significant future as a power source for electric automobiles and in space vehicles. The cell is shown in the figure below.

Electrodes are hollow tubes made of porous compressed carbon impregnated with platinum (increased surface area acts as a catalyst). The electrolyte is KOH. Hydrogen is oxidized at the anode giving electrons to the outer circuit while the electrons are accepted at the cathode where reduction occurs and in this way current flows. The cell products can be regenerated externally into fuel for return to the cell. The following cell reactions take place at the two electrodes.
2H₂  → 4H⁺ + 4e^–             (oxidation)
O₂ + 4H⁺ + 4e^– → 2H₂O  (reduction)
______________________________
2H₂ + O₂ → 2H₂O              (overall)
Fuel cells are very efficient and convert about 75% of the fuel into electricity. The major drawback of the fuel cells is that they are very costly. The gases must be of very high purity otherwise even a trace of impurity may poison the platinum which severely degrades its efficiency.