In redox reactions, electrons are transferred from one reactant to another. One reactant undergoes oxidation, losing electrons, while the other undergoes reduction, gaining electrons.
Key Concepts:
The oxidation number of an atom is a quantitative measure of its electron loss or gain in a compound. An increase in oxidation number suggests oxidation, while a decrease suggests reduction.
Oxidizing Agent: A substance that accepts electrons from another substance, causing it to be oxidized.
Reducing Agent: A substance that donates electrons to another substance, causing it to be reduced.
Half-Reactions: Reactions that show either the oxidation or reduction process separately.
Balancing Redox Reactions: Redox reactions must be balanced in terms of both mass and charge. This often involves using half-reactions and balancing the number of electrons transferred.
Types of Redox Reactions:
Combination reactions
Decomposition reactions
Displacement reactions
Disproportionation reactions
Applications of Redox Reactions:
Corrosion: The process of oxidation of metals.
Electrochemistry: Electrochemistry is the branch of chemistry that explores the connection between electrical energy and chemical processes.
Batteries: Devices that store and release electrical energy through redox reactions.
Metallurgy: The extraction of metals from their ores, often involving redox processes.
Important Points:
Redox reactions are essential for many chemical and biological processes.
Understanding redox reactions is crucial for studying various fields like chemistry, biology, and environmental science.
A mastery of balancing redox reactions is a fundamental aspect of chemical knowledge.
By understanding the concepts of oxidation, reduction, and electron transfer, you can effectively analyze and predict the outcomes of redox reactions in various contexts.
1. Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4) 2.12 H2O
Ans : Assigning Oxidation Numbers
Rules for assigning oxidation numbers:
Applying these rules to the given compounds:
(a) NaH₂PO₄:
Na: +1
H: +1
O: -2
Define the oxidation number of phosphorus as “x”.
1 + 2(1) + x + 4(-2) = 0
x = +5
(b) NaHSO₄:
Na: +1
H: +1
O: -2
Assign the symbol “x” to the oxidation number of sulfur.
1 + 1 + x + 4(-2) = 0
x = +6
(c) H₄P₂O₇:
H: +1
O: -2
denote “x” to be the oxidation state of phosphorus.
4(1) + 2x + 7(-2) = 0
x = +5
(d) K₂MnO₄:
K: +1
O: -2
Consider “x” to be the oxidation number of manganese.
2(1) + x + 4(-2) = 0
x = +7
(e) CaO₂:
Ca: +2
Consider “x” to be the oxidation state of oxygen.
+2 + 2x = 0
x = -1
(f) NaBH₄:
Na: +1
H: +1
Consider “x” to be the oxidation state of boron.
1 + 4(1) + x = 0
x = -3
(g) H₂S₂O₇:
H: +1
O: -2
Assign the symbol “x” to the oxidation number of sulfur.
2(1) + 2x + 7(-2) = 0
x = +6
(h) KAl(SO₄)₂·12H₂O:
K: +1
Al: +3
O: -2
Assign the symbol “x” to the oxidation number of sulfur.
1 + 3 + 2(x + 4(-2)) + 12(1 + (-2)) = 0
x = +6
Therefore, the underlined elements and their oxidation numbers are:
NaH₂PO₄: P = +5
NaHSO₄: S = +6
H₄P₂O₇: P = +5
K₂MnO₄: Mn = +7
CaO₂: O = -1
NaBH₄: B = -3
H₂S₂O₇: S = +6
KAl(SO₄)₂·12H₂O: S = +6
2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
Ans : Assigning Oxidation Numbers
Rules for assigning oxidation numbers:
An isolated atom has an oxidation state of zero.
The oxidation state of a single-atom ion is numerically equal to its ionic charge.
Oxidation number of hydrogen is +1, except when it is combined with a more electronegative element (e.g., in NaH, it is -1).
Oxidation number of oxygen is -2, except in peroxides (e.g., H₂O₂, where it is -1) and superoxides (e.g., KO₂, where it is -0.5).
Applying these rules to the given compounds:
(a) KI₃:
K: +1
Denote oxidation number of I be x.
1 + 3x = 0
x = -1/3
(b) H₂S₄O₆:
H: +1
O: -2
Denote oxidation number of S be x.
2(1) + 4x + 6(-2) = 0
x = +2.5
(c) Fe₃O₄:
O: -2
Denote oxidation number of Fe be x.
3x + 4(-2) = 0
x = +8/3
(d) CH₃CH₂OH:
H: +1
O: -2
Consider x to be the oxidation number of carbon.
2(1) + x + 3(1) + 1(-2) = 0
x = -2
(e) CH₃COOH:
H: +1
O: -2
Consider x to be the oxidation number of carbon.
2(1) + x + 2(1) + 2(-2) = 0
x = 0
Rationalizing the Results:
KI₃: The fractional oxidation number of iodine (-1/3) indicates that there is a resonance structure involving different oxidation states for the iodine atoms.
H₂S₄O₆: The fractional oxidation number of sulfur (+2.5) suggests that there is a combination of sulfur atoms with different oxidation states in the molecule.
Fe₃O₄: The fractional oxidation number of iron (+8/3) indicates that there is a mixture of Fe²⁺ and Fe³⁺ ions in the compound.
CH₃CH₂OH and CH₃COOH: The oxidation number of carbon in these organic molecules is -2, which is typical for carbon in organic molecules.
These results highlight the complexity of oxidation number assignments in certain compounds, especially those with multiple atoms of the same element or those with resonance structures.
3 .Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
Ans : Analyzing Redox Reactions
In a redox reaction, the oxidation numbers of the elements involved undergo a transformation.
(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)
Oxidation: H₂ → 2H⁺ (oxidation number increases from 0 to +1)
Reduction: Cu²⁺ (in CuO) → Cu (oxidation number decreases from +2 to 0)
(b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Oxidation: CO → CO₂ (oxidation number of carbon increases from +2 to +4)
Reduction: Fe³⁺ (in Fe₂O₃) → Fe (oxidation number decreases from +3 to 0)
(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
Oxidation: B (in BCl₃) → B (in B₂H₆) (oxidation number increases from +3 to -1)
Reduction: Al (in LiAlH₄) → Al (in AlCl₃) (oxidation number remains the same)
Note: While aluminum remains in the same oxidation state, the hydrogen atoms in LiAlH₄ are oxidized, and the boron atoms in BCl₃ are reduced.
(d) 2K(s) + F₂(g) → 2KF(s)
Oxidation: K → K⁺ (oxidation number increases from 0 to +1)
Reduction: F₂ → 2F⁻ (oxidation number decreases from 0 to -1)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Oxidation: N (in NH₃) → N (in NO) (oxidation number increases from -3 to +2)
Reduction: O₂ → O (in H₂O) (oxidation number decreases from 0 to -2)
Conclusion:
In all of these reactions, there is a change in oxidation numbers, indicating that they are redox reactions..
4. Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Ans :
The oxidation number (O.N.) of fluorine (F) decreases from 0 in F₂ to -1 in HF. This indicates that fluorine is gaining electrons, which is a reduction process.
At the same time, the oxidation number of fluorine increases from 0 in F₂ to +1 in HOF. This indicates that fluorine is losing electrons, which is an oxidation process.
Since fluorine undergoes both oxidation and reduction in the reaction, it is a redox reaction. More specifically, it is a disproportionation reaction, where the same element undergoes both oxidation and reduction in the same reaction.
5. Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7 2– and NO3 – . Suggest structure of these compounds. Count for the fallacy.
Ans :
Structure:
The nitrate ion exhibits a trigonal planar geometry, consisting of a central nitrogen atom bonded to three oxygen atoms.
In summary, the oxidation numbers of sulfur, chromium, and nitrogen in H₂SO₅, Cr₂O₇²⁻, and NO₃⁻ are +6, +6, and +5, respectively. The fallacy in calculating the oxidation number of sulfur in H₂SO₅ is resolved by considering its peroxy structure
6. Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
Ans : (a) Hg(II)Cl2,
(b) Ni(II)SO4,
(c)Sn(IV)O2
(d) T12(I)SO4,
(e) Fe2(III)(S04)3,
(f) Cr2(III)O3.
7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Ans :
8. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?
Ans : The oxidizing and reducing properties of a compound depend on the oxidation state of the central atom and the stability of the compound.
Sulfur dioxide (SO₂) and hydrogen peroxide (H₂O₂):
Sulfur in SO₂: The oxidation state of sulfur in SO₂ is +4. This is a relatively high oxidation state, making SO₂ a good oxidizing agent. However, sulfur can also be oxidized to a higher oxidation state (+6) in compounds like SO₃ or H₂SO₄. This makes SO₂ a potential reducing agent as well.
Oxygen in H₂O₂: The oxidation state of oxygen in H₂O₂ is -1. This is a relatively low oxidation state, making H₂O₂ a good reducing agent. However, oxygen can also be reduced to a lower oxidation state (-2) in water (H₂O). This makes H₂O₂ a potential oxidizing agent as well.
Ozone (O₃) and nitric acid (HNO₃):
Oxygen in O₃: The oxidation state of oxygen in O₃ is 0. This is a relatively high oxidation state, making O₃ a strong oxidizing agent. However, it cannot be further oxidized.
Nitrogen in HNO₃: The oxidation state of nitrogen in HNO₃ is +5. This is the highest oxidation state for nitrogen, making it a strong oxidizing agent. It cannot be further oxidized.
Therefore, ozone and nitric acid can only act as oxidizing agents because they are already in their highest possible oxidation states. Their central atoms cannot be further oxidized, limiting their ability to act as reducing agents.
9. Consider the reactions:
(a) 6 CO2(g) + 6H2O(l) → C6 H12 O6(aq) + 6O2(g)
(b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)
Why it is more appropriate to write these reactions as :
(a) 6CO2(g) + 12H2O(l) → C6 H12 O6(aq) + 6H2O(l) + 6O2(g)
(b) O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g).
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Ans : Analyzing the Reactions
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g)
Original Reaction: This reaction represents the photosynthesis process, where carbon dioxide and water are converted into glucose and oxygen.
Revised Reaction: 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) + 6O₂(g)
The revised reaction is more accurate because it explicitly shows the water molecules involved in the process. In photosynthesis, water molecules are both reactants and products. The revised equation emphasizes this fact.
(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O₂(g)
Original Reaction: This reaction represents the decomposition of ozone (O₃) by hydrogen peroxide (H₂O₂).
Revised Reaction: O₃(g) + H₂O₂(l) → H₂O(l) + O₂(g) + O₂(g)
The revised reaction is more accurate because it explicitly shows the formation of two oxygen molecules as products. This is consistent with the decomposition of ozone, which releases molecular oxygen.
Investigating the Reaction Paths
To investigate the reaction paths of these reactions, we can use isotope labeling and kinetic studies.
Isotope Labeling
Isotope labeling involves using isotopes of elements (e.g., ¹⁸O) to track the movement of atoms during a reaction.
By labeling specific atoms in the reactants and analyzing the products, we can determine the step-by-step process of the reaction.
For example, in the photosynthesis reaction, we could use ¹⁸O-labeled water to track the oxygen atoms in the products, helping us understand the mechanism of oxygen evolution.
Kinetic Studies
Kinetic studies involve measuring the rate of a reaction under different conditions (e.g., temperature, concentration).
By analyzing the rate data, we can infer the mechanism of the reaction. For example, if the rate of a reaction is first-order with respect to a reactant, it suggests that the reactant is involved in the rate-determining step.
By combining isotope labeling and kinetic studies, we can gain valuable insights into the reaction pathways of both photosynthesis and the decomposition of ozone by hydrogen peroxide.
10. The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
Ans : AgF₂ is an unstable compound due to the high oxidation state of silver (+2). Silver is generally more stable in the +1 oxidation state. The +2 oxidation state is relatively uncommon for silver, making AgF₂ unstable and prone to decomposition.
The instability of AgF₂ also contributes to its strong oxidizing properties. Compounds with high oxidation states tend to be strong oxidizing agents because they have a greater tendency to accept electrons to achieve a more stable lower oxidation state. In the case of AgF₂, the silver ion has a high oxidation state (+2), making it a strong oxidizing agent.
When AgF₂ acts as an oxidizing agent, it accepts electrons from other substances, causing those substances to be oxidized. This oxidizing behavior is often associated with the release of oxygen or the formation of higher oxidation state products.
11 .Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Ans : The Effect of Excess Reactants on Oxidation States
Statement: Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess, and the compound formed will have a higher oxidation state if there is an excess of the oxidizing agent.
The oxidation state of an element reflects its electron configuration. When an oxidizing agent (electron acceptor) is in excess, it can remove more electrons from the reducing agent, leading to a higher oxidation state in the product. Conversely, when a reducing agent (electron donor) is in excess, it can provide more electrons to the oxidizing agent, resulting in a lower oxidation state in the product.
Illustrations:
Reaction of Copper(II) Sulfate with Iron(II) Sulfate:
Balanced equation: CuSO₄(aq) + FeSO₄(aq) → Cu(s) + Fe₂(SO₄)₃(aq)
Analysis: Cu²⁺ is reduced to Cu (lower oxidation state), while Fe²⁺ is oxidized to Fe³⁺ (higher oxidation state). Since FeSO₄ is in excess, the reaction proceeds to form Fe³⁺.
Reaction of Potassium Permanganate with Oxalic Acid:
Balanced equation: 2KMnO₄(aq) + 5H₂C₂O₄(aq) + 3H₂SO₄(aq) → 2MnSO₄(aq) + K₂SO₄(aq) + 10CO₂(g) + 8H₂O(l)
Analysis: MnO₄⁻ is reduced to Mn²⁺ (lower oxidation state), while C₂O₄²⁻ is oxidized to CO₂ (higher oxidation state). If KMnO₄ is in excess, more C₂O₄²⁻ will be oxidized to CO₂.
Reaction of Iodine with Sodium Thiosulfate:
Balanced equation: I₂(aq) + 2Na₂S₂O₃(aq) → 2NaI(aq) + Na₂S₄O₆(aq)
Analysis: I₂ is reduced to I⁻ (lower oxidation state), while S₂O₃²⁻ is oxidized to S₄O₆²⁻ (higher oxidation state). If I₂ is in excess, more S₂O₃²⁻ will be oxidized to S₄O₆²⁻.
In each of these examples, the oxidation state of the product depends on the relative amounts of the oxidizing and reducing agents. When the oxidizing agent is in excess, the reducing agent is oxidized to a higher oxidation state, and vice versa.
12. How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?
Ans : (a) Use of Alkaline Potassium Permanganate in the Oxidation of Toluene
Reason:
Selectivity: Alkaline potassium permanganate (KMnO₄) is a milder oxidizing agent compared to acidic potassium permanganate. This selectivity is crucial in the oxidation of toluene to benzoic acid, as it prevents further oxidation of the benzene ring.
Reaction Conditions: The alkaline conditions provided by potassium hydroxide (KOH) in alkaline KMnO₄ favor the formation of benzoic acid.
Balanced Redox Equation:
C₆H₅CH₃(l) + 2KMnO₄(aq) + 3KOH(aq) → C₆H₅COO⁻(aq) + 2MnO₂(s) + 2K₂CO₃(aq) + 4H₂O(l)
In this reaction, toluene (C₆H₅CH₃) is oxidized to benzoate ion (C₆H₅COO⁻), while KMnO₄ is reduced to MnO₂.
(b) Reaction of Concentrated H₂SO₄ with Chlorides and Bromides
Reason:
Strength of Reducing Agent: The halide ions (Cl⁻ and Br⁻) have different reducing strengths. Iodide (I⁻) is a stronger reducing agent than bromide (Br⁻), and bromide is stronger than chloride (Cl⁻).
Oxidizing Power of H₂SO₄: Concentrated H₂SO₄ can act as a strong oxidizing agent, especially when heated.
Reactions:
With Chloride (Cl⁻):
H₂SO₄(aq) + NaCl(s) → NaHSO₄(aq) + HCl(g)
The colorless gas HCl is produced.
With Bromide (Br⁻):
2H₂SO₄(aq) + 2NaBr(s) → Na₂SO₄(aq) + Br₂(l) + SO₂(g) + 2H₂O(l)
The red vapor of bromine (Br₂) is produced along with other products.
Explanation:
When concentrated H₂SO₄ reacts with chlorides, it oxidizes Cl⁻ to HCl, a colorless gas.
However, when H₂SO₄ reacts with bromides, it is able to oxidize Br⁻ to Br₂, a reddish-brown liquid or gas. This is because Br⁻ is a stronger reducing agent than Cl⁻, and H₂SO₄ has sufficient oxidizing power to oxidize it.
13. Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
(b) HCHO(l) + 2[Ag (NH3) 2] +(aq) + 3OH– (aq) → 2Ag(s) + HCOO– (aq) + 4NH3(aq) + 2H2O(l)
(c) HCHO (l) + 2 Cu2+(aq) + 5 OH– (aq) → Cu2O(s) + HCOO– (aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Ans :
14. Consider the reactions :
2 S2O3 2– (aq) + I2(s) → S4 O6 2–(aq) + 2I– (aq)
S2O3 2–(aq) + 2Br2(l) + 5 H2O(l) → 2SO4 2–(aq) + 4Br– (aq) + 10H+(aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine ?
Ans : The average oxidation number (O.N.) of sulfur (S) in S₂O₃²⁻ is +2, while in S₄O₆²⁻ it is +2.5. The O.N. of sulfur in SO₄²⁻ is +6.
Since Br₂ is a stronger oxidizing agent than I₂, it can oxidize the sulfur in S₂O₃²⁻ to a higher oxidation state of +6, forming SO₄²⁻ ions. On the other hand, I₂, being a weaker oxidizing agent, can only oxidize the sulfur in S₂O₃²⁻ to a lower oxidation state of +2.5, forming S₄O₆²⁻ ions.
This difference in oxidizing power between Br₂ and I₂ explains why thiosulfate (S₂O₃²⁻) reacts differently with these two halogens.
15 .Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Ans : The halogens have a strong tendency to gain electrons, making them strong oxidizing agents. Their relative oxidizing power is measured by their electrode potentials. Fluorine (F₂) has the highest electrode potential (+2.87 V), indicating that it is the strongest oxidizing agent among the halogens.
The oxidizing strength of the halogens diminishes in the sequence:
I₂< Br₂ < Cl₂ < F₂
This is evident from the following reactions:
F₂ can oxidize Cl⁻ to Cl₂, Br⁻ to Br₂, and I⁻ to I₂.
Cl₂ can oxidize Br⁻ to Br₂ and I⁻ to I₂, but not F⁻ to F₂.
Br₂ can oxidize I⁻ to I₂, but not F⁻ to F₂ or Cl⁻ to Cl₂.
Conversely, halide ions (F⁻, Cl⁻, Br⁻, I⁻) have a tendency to lose electrons, making them reducing agents.
Their reducing power increases in the order:
I⁻ < Br⁻ < Cl⁻ < F⁻
This is supported by the following reactions:
HI and HBr can reduce H₂SO₄ to SO₂, while HCl and HF cannot.
F⁻ cannot reduce Cu²⁺ to Cu⁺, while Br⁻ can.
Therefore, the reducing character of hydrohalic acids decreases in the order:
HI > HBr > HCl > HF
16. Why does the following reaction occur ?
XeO6 4– (aq) + 2F– (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l)
What conclusion about the compound Na4XeO6 (of which XeO6 4– is a part) can be drawn from the reaction.
Ans : The given reaction occurs because XeO₆⁴⁻ is a strong oxidizing agent. It can oxidize fluoride ions (F⁻) to fluorine gas (F₂) while itself being reduced to xenon trioxide (XeO₃).
Here’s a breakdown of the reaction:
Oxidation: 2F⁻(aq) → F₂(g) + 2e⁻
Reduction: XeO₆⁴⁻(aq) + 6H⁺(aq) + 2e⁻ → XeO₃(g) + 3H₂O(l)
Overall reaction: XeO₆⁴⁻(aq) + 2F⁻(aq) + 6H⁺(aq) → XeO₃(g) + F₂(g) + 3H₂O(l)
Conclusion about Na₄XeO₆:
Strong Oxidizing Agent: Since XeO₆⁴⁻ can oxidize F⁻ to F₂, it is a strong oxidizing agent.
Unstable Compound: The tendency of XeO₆⁴⁻ to undergo reduction suggests that it is a relatively unstable compound.
High Oxygen Content: The high oxidation state of xenon in XeO₆⁴⁻ (Xe is in the +8 oxidation state) makes it a strong oxidizing agent.
Therefore, Na₄XeO₆ is a strong oxidizing agent due to the high oxidation state of xenon and the instability of the XeO₆⁴⁻ ion.
17. Consider the reactions:
(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)
(c) C6H5CHO(l) + 2[Ag (NH3) 2] +(aq) + 3OH– (aq) → C6H5COO– (aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l)
(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH– (aq) → No change observed.
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions ?
Ans : The given reactions involve the reduction of silver ions (Ag⁺) and copper ions (Cu²⁺) by hypophosphorous acid (H₃PO₂). Based on the observed reactions, we can draw the following inferences about the behavior of Ag⁺ and Cu²⁺:
1. Ag⁺ is a stronger oxidizing agent than Cu²⁺.
In reactions (a) and (c), H₃PO₂ reduces Ag⁺ to Ag.
In reaction (d), H₃PO₂ fails to reduce Cu²⁺.
This indicates that Ag⁺ has a higher tendency to gain electrons compared to Cu²⁺, making it a stronger oxidizing agent.
2. The reducing power of H₃PO₂ is sufficient to reduce Ag⁺ but not Cu²⁺.
H₃PO₂ successfully reduces Ag⁺ to Ag in reactions (a) and (c).
However, it cannot reduce Cu²⁺ to Cu in reaction (d).
This suggests that H₃PO₂ has a moderate reducing power that is sufficient to reduce Ag⁺ but not strong enough to reduce Cu²⁺.
In summary, the reactions demonstrate that Ag⁺ is a stronger oxidizing agent than Cu²⁺, and H₃PO₂ has a reducing power that is sufficient to reduce Ag⁺ but not Cu²⁺.
18. Balance the following redox reactions by ion – electron method :
(a) MnO4 – (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)
(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4 – (aq) (in acidic solution)
(c) H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO4 2– (aq) (in acidic solution)
Ans : Balancing Redox Reactions by Ion-Electron Method
(a) MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s) (basic medium)
Half-reactions:
Oxidation: 2I⁻(aq) → I₂(s)
Reduction: MnO₄⁻(aq) → MnO₂(s)
Balance atoms other than O and H:
Oxidation: 2I⁻(aq) → I₂(s)
Reduction: MnO₄⁻(aq) → MnO₂(s)
Balance oxygen with H₂O:
Oxidation: 2I⁻(aq) → I₂(s)
Reduction: MnO₄⁻(aq) + 2H₂O(l) → MnO₂(s) + 4OH⁻(aq)
Balance hydrogen with OH⁻:
Oxidation: 2I⁻(aq) → I₂(s)
Reduction: MnO₄⁻(aq) + 2H₂O(l) + 3e⁻ → MnO₂(s) + 4OH⁻(aq)
Balance charge with electrons:
Oxidation: 2I⁻(aq) → I₂(s) + 2e⁻
Reduction: MnO₄⁻(aq) + 2H₂O(l) + 3e⁻ → MnO₂(s) + 4OH⁻(aq)
Multiply half-reactions to balance electrons:
Oxidation: 3(2I⁻(aq) → I₂(s) + 2e⁻)
Reduction: 2(MnO₄⁻(aq) + 2H₂O(l) + 3e⁻ → MnO₂(s) + 4OH⁻(aq))
Add half-reactions and simplify:
6I⁻(aq) + 2MnO₄⁻(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH⁻(aq)
Balanced Equation:
6I⁻(aq) + 2MnO₄⁻(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH⁻(aq)
(b) MnO₄⁻(aq) + SO₂(g) → Mn²⁺(aq) + HSO₄⁻(aq) (acidic solution)
Half-reactions:
Oxidation: SO₂(g) → HSO₄⁻(aq)
Reduction: MnO₄⁻(aq) → Mn²⁺(aq)
Balance atoms other than O and H:
Oxidation: SO₂(g) → HSO₄⁻(aq)
Reduction: MnO₄⁻(aq) → Mn²⁺(aq)
Balance oxygen with H₂O:
Oxidation: SO₂(g) + 2H₂O(l) → HSO₄⁻(aq) + 3H⁺(aq)
Reduction: MnO₄⁻(aq) + 4H⁺(aq) → Mn²⁺(aq) + 2H₂O(l)
Balance hydrogen with H⁺:
Oxidation: SO₂(g) + 2H₂O(l) → HSO₄⁻(aq) + 3H⁺(aq)
Reduction: MnO₄⁻(aq) + 4H⁺(aq) → Mn²⁺(aq) + 2H₂O(l)
Balance charge with electrons:
Oxidation: SO₂(g) + 2H₂O(l) → HSO₄⁻(aq) + 3H⁺(aq) + 2e⁻
Reduction: MnO₄⁻(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Multiply half-reactions to balance electrons:
Oxidation: 5SO₂(g) + 10H₂O(l) → 5HSO₄⁻(aq) + 15H⁺(aq) + 10e⁻
Reduction: 2MnO₄⁻(aq) + 10e⁻ → 2Mn²⁺(aq) + 8H₂O(l)
Add half-reactions and simplify:
5SO₂(g) + 2MnO₄⁻(aq) + 2H₂O(l) → 5HSO₄⁻(aq) + 2Mn²⁺(aq) + 2H⁺(aq)
Balanced Equation:
5SO₂(g) + 2MnO₄⁻(aq) + 2H₂O(l) → 5HSO₄⁻(aq) + 2Mn²⁺(aq) + 2H⁺(aq)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (acidic solution)
Half-reactions:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq)
Reduction: H₂O₂(aq) → H₂O(l)
Balance atoms other than O and H:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq)
Reduction: H₂O₂(aq) → H₂O(l)
Balance oxygen with H₂O:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq)
Reduction: H₂O₂(aq) → 2H₂O(l)
Balance hydrogen with H⁺:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq)
Reduction: H₂O₂(aq) + 2H⁺(aq) → 2H₂O(l)
Balance charge with electrons:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Reduction: H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l)
Multiply half-reactions to balance electrons:
Oxidation: 2Fe²⁺(aq) → 2Fe³⁺(aq) + 2e⁻
Reduction: H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l)
Add half-reactions and simplify:
2Fe²⁺(aq) + H₂O₂(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l)
Balanced Equation:
2Fe²⁺(aq) + H₂O₂(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l)
(d) Cr₂O₇²⁻(aq) + SO₂(g) → Cr³⁺(aq) + SO₄²⁻(aq) (acidic solution)
Half-reactions:
Oxidation: SO₂(g) → SO₄²⁻(aq)
Reduction: Cr₂O₇²⁻(aq) → Cr³⁺(aq)
Balance atoms other than O and H:
Oxidation: SO₂(g) → SO₄²⁻(aq)
Reduction: Cr₂O₇²⁻(aq) → 2Cr³⁺(aq)
Balance oxygen with H₂O:
Oxidation: SO₂(g) + 2H₂O(l) → SO₄²⁻(aq) + 4H⁺(aq)
Reduction: Cr₂O₇²⁻(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l)
Balance hydrogen with H⁺:
Oxidation: SO₂(g) + 2H₂O(l) → SO₄²⁻(aq) + 4H⁺(aq)
Reduction: Cr₂O₇²⁻(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l)
Balance charge with electrons:
Oxidation: SO₂(g) + 2H₂O(l) → SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻
Reduction: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
Multiply half-reactions to balance electrons:
Oxidation: 3SO
19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) P4(s) + OH– (aq) → PH3(g) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl– (g)
(c) Cl2O7 (g) + H2O2(aq) → ClO2 – (aq) + O2(g) + H+
Ans : Balancing Redox Reactions in Basic Medium
(a) P₄(s) + OH⁻(aq) → PH₃(g) + HPO₂⁻(aq)
Ion-Electron Method:
Half-reactions:
Oxidation: P₄(s) → 4HPO₂⁻(aq)
Reduction: P₄(s) → 4PH₃(g)
Balance atoms other than O and H:
Oxidation: P₄(s) → 4HPO₂⁻(aq)
Reduction: P₄(s) → 4PH₃(g)
Balance oxygen with H₂O
Oxidation: P₄(s) + 16H₂O(l) → 4HPO₂⁻(aq) + 12H⁺(aq)
Reduction: P₄(s) + 12H⁺(aq) → 4PH₃(g) + 12H₂O(l)
Balance hydrogen with OH⁻:
Oxidation: P₄(s) + 16H₂O(l) + 12OH⁻(aq) → 4HPO₂⁻(aq) + 20H⁺(aq)
Reduction: P₄(s) + 12H⁺(aq) + 12OH⁻(aq) → 4PH₃(g) + 12H₂O(l)
Balance charge with electrons:
Oxidation: P₄(s) + 16H₂O(l) + 12OH⁻(aq) → 4HPO₂⁻(aq) + 20H⁺(aq) + 20e⁻
Reduction: P₄(s) + 12H⁺(aq) + 12OH⁻(aq) + 12e⁻ → 4PH₃(g) + 12H₂O(l)
Multiply half-reactions to balance electrons:
Oxidation: 3P₄(s) + 48H₂O(l) + 36OH⁻(aq) → 12HPO₂⁻(aq) + 60H⁺(aq) + 60e⁻
Reduction: P₄(s) + 12H⁺(aq) + 12OH⁻(aq) + 12e⁻ → 4PH₃(g) + 12H₂O(l)
Add half-reactions and simplify:
4P₄(s) + 36H₂O(l) + 24OH⁻(aq) → 12HPO₂⁻(aq) + 48H⁺(aq) + 48e⁻
Balanced Equation:
4P₄(s) + 12OH⁻(aq) → 12HPO₂⁻(aq) + 4PH₃(g)
Oxidation Number Method:
Assign oxidation numbers: P in P₄ = 0, P in HPO₂⁻ = +5
Identify redox: P is both oxidized and reduced.
Balance P atoms: 4P₄(s) → 12HPO₂⁻(aq) + 4PH₃(g)
Balance O and H with H₂O and OH⁻:
4P₄(s) + 16H₂O(l) + 12OH⁻(aq) → 12HPO₂⁻(aq) + 20H⁺(aq)
Balance charge with electrons:
4P₄(s) + 16H₂O(l) + 12OH⁻(aq) → 12HPO₂⁻(aq) + 20H⁺(aq) + 20e⁻
Simplify:
4P₄(s) + 12OH⁻(aq) → 12HPO₂⁻(aq) + 4PH₃(g)
Oxidizing Agent: P₄ (reduced)
Reducing Agent: P₄ (oxidized)
(b) N₂H₄(l) + ClO₃⁻(aq) → NO(g) + Cl⁻(g)
Ion-Electron Method:
Half-reactions:
Oxidation: N₂H₄(l) → 2NO(g)
Reduction: ClO₃⁻(aq) → Cl⁻(g)
Balance atoms other than O and H:
Oxidation: N₂H₄(l) → 2NO(g)
Reduction: ClO₃⁻(aq) → Cl⁻(g)
Balance oxygen with H₂O:
Oxidation: N₂H₄(l) + 2H₂O(l) → 2NO(g) + 4H⁺(aq)
Reduction: ClO₃⁻(aq) + 6H⁺(aq) → Cl⁻(g) + 3H₂O(l)
Balance hydrogen with OH⁻:
Oxidation: N₂H₄(l) + 2H₂O(l) + 4OH⁻(aq) → 2NO(g) + 8H⁺(aq) + 4OH⁻(aq)
Reduction: ClO₃⁻(aq) + 6H⁺(aq) + 6OH⁻(aq) → Cl⁻(g) + 3H₂O(l) + 6OH⁻(aq)
Balance charge with electrons:
Oxidation: N₂H₄(l) + 2H₂O(l) + 4OH⁻(aq) → 2NO(g) + 8H⁺(aq) + 4OH⁻(aq) + 8e⁻
Reduction: ClO₃⁻(aq) + 6H⁺(aq) + 6OH⁻(aq) + 6e⁻ → Cl⁻(g) + 3H₂O(l) + 6OH⁻(aq)
Multiply half-reactions to balance electrons:
Oxidation: 3N₂H₄(l) + 6H₂O(l) + 12OH⁻(aq) → 6NO(g) + 24H⁺(aq) + 12OH⁻(aq) + 24e⁻
Reduction: 2ClO₃⁻(aq) + 12H⁺(aq) + 12OH⁻(aq) + 12e⁻ → 2Cl⁻(g) + 6H₂O(l) + 12OH⁻(aq)
Add half-reactions and simplify:
3N₂H₄(l) + 2ClO₃⁻(aq) → 6NO(g) + 2Cl⁻(g) + 6H₂O(l)
Balanced Equation:
3N₂H₄(l) + 2ClO₃⁻(aq) → 6NO(g) + 2Cl⁻(g) + 6H₂O(l)
Oxidation Number Method:
Assign oxidation numbers: N in N₂H₄ = -2, N in NO = +2, Cl in ClO₃⁻ = +5, Cl in Cl⁻ = -1
Identify redox: N is oxidized, Cl is reduced.
Balance atoms: 3N₂H₄(l) + 2ClO₃⁻(aq) → 6NO(g) + 2Cl⁻(g)
Balance O and H with H₂O and OH⁻:
3N₂H₄(l) + 2ClO₃⁻(aq) + 6H₂O(l) → 6NO(g) + 2Cl⁻(g) + 12H⁺(aq)
Balance charge with electrons:
3N₂H₄(l) + 2ClO₃⁻(aq) + 6H₂O(l) + 12e⁻ → 6NO(g) + 2Cl⁻(g) + 12H⁺(aq) + 12e⁻
Simplify:
3N₂H₄(l) + 2ClO₃⁻(aq) → 6NO(g) + 2Cl⁻(g) + 6H₂O(l)
Oxidizing Agent: ClO₃⁻ (reduced)
Reducing Agent: N₂H₄ (oxidized)
(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂⁻(aq) + O₂(g) + H⁺
Ion-Electron Method:
Half-reactions:
Oxidation: H₂O₂(aq) → O₂(g)
Reduction: Cl₂O₇(g) → ClO₂⁻
20. What sorts of informations can you draw from the following reaction ?
(CN)2(g) + 2OH– (aq) → CN– (aq) + CNO– (aq) + H2O(l)
Ans : The given reaction represents the disproportionation of cyanogen (CN₂) in an alkaline medium. Here’s what we can infer from the equation:
1. Redox Reaction: It’s a redox reaction as the oxidation state of the central atom (carbon) changes. In CN₂, carbon has an oxidation state of +2. In CN⁻, it’s -3, and in CNO⁻, it’s +2.
2. Products: The reaction produces two different products: cyanide ion (CN⁻) and cyanate ion (CNO⁻). This indicates that cyanogen undergoes both oxidation and reduction simultaneously.
3. Alkaline Medium: The presence of hydroxide ions (OH⁻) suggests that the reaction occurs in an alkaline environment.
4. Water Formation: The formation of water (H₂O) indicates that a proton transfer is involved in the reaction.
5. Equilibrium: The reaction might be reversible, depending on the conditions. If the reaction reaches equilibrium, the concentrations of the products and reactants will remain constant.
Overall, the reaction shows that cyanogen can undergo disproportionation in an alkaline medium to produce cyanide and cyanate ions.
21. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.
Ans : The skeletal equation for the disproportionation reaction of Mn³⁺ is:
Mn³⁺(aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)
To balance the oxidation half-reaction:
Balance the oxidation numbers by adding an electron to the right side:
Mn³⁺(aq) → MnO₂(s) + e⁻
Balance the charge by adding 4H⁺ ions to the right side:
Mn³⁺(aq) → MnO₂(s) + 4H⁺(aq) + e⁻
Balance the oxygen atoms by adding 2H₂O molecules to the left side:
Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺(aq) + e⁻
To balance the reduction half-reaction:
Balance the oxidation numbers by adding an electron to the left side:
Mn³⁺(aq) + e⁻ → Mn²⁺(aq)
Adding the balanced oxidation and reduction half-reactions, we get the balanced equation for the disproportionation reaction:
2Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + Mn²⁺(aq) + 4H⁺(aq)
22. Consider the elements : Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only postive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Ans : (a) Fluorine (F) is the element that exhibits only a negative oxidation state. As the most electronegative element, it always attracts electrons in a compound, resulting in a -1 oxidation state.
(b) Cesium (Cs) is the element that exhibits only a positive oxidation state. As an alkali metal, it readily loses its valence electron to form a +1 cation.
(c) Iodine (I) exhibits both positive and negative oxidation states. It can form compounds with oxidation states ranging from -1 to +7.
(d) Neon (Ne) is the element that exhibits neither a negative nor a positive oxidation state. As a noble gas, it has a full valence shell and is generally unreactive, maintaining a neutral state.
23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Ans :
24. Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Ans :
25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?
Ans : The first step in the Ostwald process involves the oxidation of ammonia gas by oxygen gas according to the following balanced equation:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
To determine the limiting reactant and the maximum amount of nitric oxide (NO) that can be produced, we need to calculate the moles of each reactant and compare them based on the stoichiometric coefficients of the balanced equation.
Step 1:
Find moles of NH₃ and O₂:
Moles of NH₃ = mass of NH₃ / molar mass of NH₃ = 10.00 g / 17.03 g/mol ≈ 0.587 mol
Moles of O₂ = mass of O₂ / molar mass of O₂ = 20.00 g / 32.00 g/mol ≈ 0.625 mol
Step 2: Determine the limiting reactant:
The balanced equation reveals a 4:5 molar ratio between NH₃ and O₂ in the reaction. Therefore, the limiting reactant is the one that has the smallest mole ratio to its stoichiometric coefficient.
Mole ratio of NH₃ to O₂ = 0.587 mol / 4 = 0.147
Mole ratio of O₂ to NH₃ = 0.625 mol / 5 = 0.125
Since the mole ratio of O₂ to NH₃ is smaller, O₂ is the limiting reactant.
Step 3: Calculate the maximum moles of NO produced:
Using the mole ratio of O₂ to NO from the balanced equation (5 moles of O₂ produce 4 moles of NO):
Moles of NO = (0.625 mol O₂) * (4 mol NO / 5 mol O₂) = 0.5 mol NO
Step 4: Convert moles of NO to grams:
Mass of NO = moles of NO * molar mass of NO = 0.5 mol * 30.01 g/mol = 15.005 g
Therefore, the maximum weight of nitric oxide that can be obtained starting with 10.00 g of ammonia and 20.00 g of oxygen is 15.005 grams.
26. Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:
(a) Fe3+(aq) and I– (aq)
(b) Ag+(aq) and Cu(s)
(c) Fe3+ (aq) and Cu(s)
(d) Ag(s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq).
Ans : To predict whether a redox reaction is feasible, we need to compare the standard reduction potentials of the half-reactions involved. If the standard reduction potential of the oxidizing agent is more positive than that of the reducing agent, the reaction is feasible.
Here are the standard reduction potentials for the relevant species:
Fe³⁺/Fe: -0.036 V
I₂/I⁻: 0.54 V
Ag⁺/Ag: 0.80 V
Cu²⁺/Cu: 0.34 V
Br₂/Br⁻: 1.09 V
Now, let’s analyze each reaction:
(a) Fe³⁺(aq) and I⁻(aq):
Fe³⁺ is reduced to Fe²⁺: Fe³⁺ + e⁻ → Fe²⁺ (E° = -0.036 V)
I⁻ is oxidized to I₂: 2I⁻ → I₂ + 2e⁻ (E° = 0.54 V)
Net reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
E°cell = E°(cathode) – E°(anode) = 0.54 V – (-0.036 V) = 0.576 V
Since E°cell is positive, the reaction is feasible.
(b) Ag⁺(aq) and Cu(s):
Ag⁺ is reduced to Ag: Ag⁺ + e⁻ → Ag (E° = 0.80 V)
Cu is oxidized to Cu²⁺: Cu → Cu²⁺ + 2e⁻ (E° = -0.34 V)
Net reaction: 2Ag⁺ + Cu → 2Ag + Cu²⁺
E°cell = 0.80 V – (-0.34 V) = 1.14 V
Since E°cell is positive, the reaction is feasible.
(c) Fe³⁺(aq) and Cu(s):
Fe³⁺ is reduced to Fe²⁺: Fe³⁺ + e⁻ → Fe²⁺ (E° = -0.036 V)
Cu is oxidized to Cu²⁺: Cu → Cu²⁺ + 2e⁻ (E° = -0.34 V)
Net reaction: 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺
E°cell = -0.34 V – (-0.036 V) = -0.304 V
Since E°cell is negative, the reaction is not feasible.
(d) Ag(s) and Fe³⁺(aq):
Ag is oxidized to Ag⁺: Ag → Ag⁺ + e⁻ (E° = -0.80 V)
Fe³⁺ is reduced to Fe²⁺: Fe³⁺ + e⁻ → Fe²⁺ (E° = -0.036 V)
Net reaction: Ag + Fe³⁺ → Ag⁺ + Fe²⁺
E°cell = -0.036 V – (-0.80 V) = 0.764 V
Since E°cell is positive, the reaction is feasible.
(e) Br₂(aq) and Fe²⁺(aq):
Br₂ is reduced to Br⁻: Br₂ + 2e⁻ → 2Br⁻ (E° = 1.09 V)
Fe²⁺ is oxidized to Fe³⁺: Fe²⁺ → Fe³⁺ + e⁻ (E° = -0.036 V)
Net reaction: Br₂ + 2Fe²⁺ → 2Br⁻ + 2Fe³⁺
E°cell = 1.09 V – (-0.036 V) = 1.126 V
Since E°cell is positive, the reaction is feasible.
27. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution AgNO3 with platinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Ans : Predicting Electrolysis Products
Electrolysis is the process of decomposing a compound into its constituent elements using an electric current. The products of electrolysis depend on the nature of the electrolyte and the electrodes used.
(i) At the cathode: Ag⁺ ions will be reduced to Ag metal.
At the anode: Ag metal will be oxidized to Ag⁺ ions.
Net reaction: No net reaction occurs. The silver electrodes will simply dissolve and redeposit.
(ii) At the cathode: Ag⁺ ions will be reduced to Ag metal.
At the anode: Water molecules will be oxidized to oxygen gas and hydrogen ions.
Net reaction: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
(iii) At the cathode: H⁺ ions will be reduced to hydrogen gas.
At the anode: Water molecules will be oxidized to oxygen gas and hydrogen ions.
Net reaction: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
(iv) At the cathode: Cu²⁺ ions will be reduced to Cu metal.
At the anode : Chloride ions will lose electrons at the anode, forming chlorine gas.
Net reaction: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Summary:
Silver electrodes with AgNO₃: No net reaction.
Platinum electrodes with AgNO₃: Oxygen gas and hydrogen ions are produced.
Platinum electrodes with dilute H₂SO₄: Hydrogen gas and oxygen gas are produced.
Platinum electrodes with CuCl₂: Copper metal and chlorine gas are produced.
28. Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Ans : The metals can be arranged in the following order of their ability to displace each other from solutions of their salts:
Mg > Al > Zn > Fe > Cu
This order is based on their reactivity or reducing power. A metal with a higher reactivity can displace a metal with a lower reactivity from its salt solution.
Here’s a breakdown of why this order is correct:
Mg: Magnesium is the most reactive metal in this list and can displace all the others.
Al: Aluminum can displace all the metals except magnesium.
Zn: Zinc can displace iron and copper, but not magnesium or aluminium.
Fe: Iron can only displace copper.
Cu: Copper is the least reactive metal and cannot displace any of the others.
29. Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37V. Cr3+/Cr = –0.74V arrange these metals in their increasing order of reducing power.
Ans : A metal with a more negative standard reduction potential is a stronger reducing agent. This means it is more likely to lose electrons and reduce another species.
Given the standard electrode potentials:
K⁺/K = -2.93 V
Ag⁺/Ag = 0.80 V
Hg₂²⁺/Hg = 0.79 V
Mg²⁺/Mg = -2.37 V
Cr³⁺/Cr = -0.74 V
Arranging the metals in increasing order of reducing power:
Ag (least reducing power)
Hg
Cr
Mg
K (most reducing power)
Therefore, the increasing order of reducing power for the given metals is Ag < Hg < Cr < Mg < K.
30. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) +2Ag(s) takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Ans : Galvanic Cell for Zn-Ag Reaction
Cell Diagram:
Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)
Components:
Anode: Zn(s) electrode (negative)
Cathode: Ag(s) electrode (positive)
Electrolyte: Solutions of Zn²⁺ and Ag⁺ ions
Salt bridge: A porous barrier containing a neutral electrolyte (e.g., KCl) to maintain electrical neutrality
Reactions:
Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
Flow of Electrons and Ions:
Electrons: The external circuit carries electrons from the zinc electrode to the silver electrode.
Ions: Zn²⁺ ions move from the anode compartment to the cathode compartment through the salt bridge. There is a movement of Ag⁺ ions from the cathode compartment to the anode compartment.
Summary:
The Zn electrode is negatively charged.
The carriers of the current in the cell are electrons in the external circuit and ions in the electrolyte.
The individual reactions occurring at each electrode are:
Anode: Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
