Chapter 7 Electrochemistry Chemistry 9 class Notes short, long question, numerical
Chapter 7 Electrochemistry Chemistry 9 class Notes short, long question numerical These 9th class chemistry notes prepared according to the syllabus of all KPK Mardan board. Other boards other than Punjab do not follow class 9 chemistry notes.
Short Questions Chapter 7 Electrochemistry Chemistry 9
Table of Contents
Q.1) Indicate which element is reduced in the following reaction:
a) Ca(s) + Br2(g) → CaBr2(s) b) 8H+ + MnO–4 + 5e– → Mn2+ + 4H2O
Answer: a) Ca(s) + Br2(g) → CaBr2(s) There is a decrease in oxidation number of Br2, from 0 to -1. So Br2 is reduced. b) 8H+ + MnO–4 + 5e– → Mn2+ + 4H2O There is a decrease in oxidation number of Mn, from +7 to +2. So Mn in reduced.
Q.2) What is the oxidation number of silver on each side of the following equation?
heat 4Ag(s) + O2(g) → 2Ag2O(s)
Answer: For the left side of equation: The oxidation number of all elements in the free state is zero. Therefore on the left side of the equation, oxidation state of silver is zero. For the right side of equation: 2[O.N of Ag] + [O.N of O] = 0 2[O.N of Ag] – 2 = 0 O.N of Ag = 2 / 2 = +1
Q.3) Why NaOH is a strong but NH4OH is a weak electrolyte?
Answer: NaOH is a strong electrolyte because it dissociates or ionizes completely in aqueous solution or in molten state and conduct electric current to a large extent. NaCl(aq) → Na+(aq) + Cl–(aq) NH4OH is a weak electrolyte because it dissociates or ionizes slightly in aqueous solution or in molten state and conduct electricity to a very small extent. NH4OH(aq) ⇋ NH4+(aq) + OH–(aq)
Q.4) How to prevent corrosion? Enlist few of the methods.
Answer: Corrosion can be prevented by a number of methods depending on the circumstances of the corroded metal. Corrosion prevention techniques are generally classified into six groups. These techniques are following: 1. Removal of stains 2. Paints and coatings 3. Alloying 4. Metallic coating or plating 5. Corrosion inhibitors 6. Cathodic protection
Q.5) Write chemical reactions that occur in Nelson’s cell.
Answer: The saturated brine solution ionizes as follow; 2NaCl→ 2Na+ + 2Cl– 2H2O → 2H++ 2OH– At anode (oxidation) 2Cl– → 2Cl + 2e– Cl + Cl →Cl2 At cathode (reduction) 2H+ + 2e– → 2H H + H → H2 2Na+ + 2OH– → 2NaOH Overall reaction 2NaCl + 2H2O → H2 + Cl2 + 2NaOH
Q.6) Write one example from daily life which involves the oxidation-reduction reaction.
Answer: Combustion of organic material in wood is an oxidation-reduction reaction. The carbon in the burning compound combines with oxygen atoms in the air and some oxygen combined with hydrogen in the compound. Therefore, the compound being burned is oxidized and the oxygen is reduced.
Q.7) Assign oxidation numbers to each atom in the following compounds.
a) HI b) PBr3 c) CaCO3 d) H3PO4 e) As3O5 f) H2SO4
Answer: a) HI O.N of H = +1 O.N of I = -1 b) PBr3 O.N of Br = -1 O.N of P + 3[O.N of Br] = 0 O.N of P +3(-1) = 0 O.N of P = +3 c) CaCO3 O.N of Ca = +2 O.N of O = -2 [O.N of Ca] + [O.N of C] + 3[O.N of O] =0 +2 + [O.N of C] + 3(-2) = 0 +2 + [O.N of C] – 6 = 0 O.N of C = +4 d) H3PO4 O.N of H = +1 O.N of O = -2 3[O.N of H] + [O.N of P] + 4[O.N of O] = 0 3(+1) + [O.N of P] + 4(-2) = 0 3 + [O.N of P] – 8 = 0 O.N of P = +5 e) As3O5 O.N of O = -2 3[O.N of As] + 5[-2] = 0 3[O.N of As] + (-10) = 0 3[O.N of As] = +10 [O.N of As] = +10/3 (Note: In Some compounds, elements do have fractional oxidation state) f) H2SO4 O.N of H = +1 O.N of O = -2 2[O.N of H] + [O.N of S] + 4[O.N of O] = 0 2(+1) + [O.N of S] + 4(-2) = 0 2 + [O.N of S] -8 = 0 O.N of S = +6
Q.8) Why O2 is necessary for rusting?
Answer: Stains and dents on the iron surface provide the site for rusting. This region is called anodic region. Iron dissolves in water to give Fe+2 and release electrons as follows: Fe → Fe+2 + 2e– The electrons liberated from the anode portion of the iron flow to the cathode. Cathode is the region with relatively high concentration of O2 near the surface surrounded by water. The free oxygen and dissolved iron bond into the salt Fe2O3.nH2O, which is called rust. 2Fe+2 + ½ O2 + (2+n)H2O → Fe2O3.nH2O + 4H+ This shows that O2 is necessary for rusting.
Q.9) Sketch the Daniel cell, labelling the cathode, anode and the direction of flow of the electrons.
Q.10) Write down some possible uses of an electrolytic cell.
Answer: (i) Many elements are made by electrolysis. For example, aluminium, zinc, sodium, hydrogen, chlorine etc. (ii) Electricity can be used to generate hydrogen and oxygen from water. (iii) Electrolytic cells can be used to purify metals (e.g Refining of Copper). (iv) Electrolytic cell can be used to generate sodium metal and chlorine gas from sodium chloride salts using Down’s cell. (v) Nelson cell is used for the preparation of Sodium hydroxide (NaOH) by electrolysis of aqueous Sodium Chloride solution. (vi) Electrolytic cells are used to electroplate metals such as tin, silver on steel.
Long Questions Chemistry Notes 9th Class
Q.1 a) What is electroplating?
Answer: Electroplating An electrolytic process inwhich a thinlayer of one metal is deposited on another metal surface. Metal ion is reduced and a solid metal is deposited on a surface is called electroplating. There are also specific types of electroplating such as copper plating, silver plating, and chromium plating. The object to be plated is made cathode (negative electrode) and the metal to be deposited is made anode (positive electrode). The electrolysis is carried out in aqueous solution of a salt of the metal being deposited.
Q.1 b) Distinguish between the nature of the anode and the cathode in such a process.
Answer: Nature of Anode: The anode is made of the metal, which is to be deposited like Cr, Ni. Anode is connected to positive terminal of battery and oxidation takes place on it. When the electric current is passed, the metal from anode converts into ions in solution and these ions are deposited on the cathode (object). Nature of Cathode: The cathode is made up of the object that is to be electroplated like some sheet made up of iron. Cathode is connected to negative terminal of battery. Reduction takes place on cathode. Metal ions coming from anode are reduced and coated on object (to be plated). As a result of this deposition, a thin layer of metal is deposited on the object (cathode).
Q.2) Differentiate between the process of oxidation and reduction. Write an equation to illustrate each.
It is the gain of oxygen atoms by an element e.g. oxidation of C into CO2 by combustion. C + O2 → CO2
It is the gain of hydrogen atoms by an element e.g. N2 molecule gains hydrogen and forms NH3. N2 + 3H2 → 2NH3
It is the loss of hydrogen atoms by an element e.g. Ammonia (NH3) oxidized to nitrogen by losing hydrogen. NH3 + 2Cl2 → N2 + 6HCl
It is the loss of oxygen atoms by an element e.g. HgO loses oxygen and form Hg. 2HgO → Hg + O2
According to the modern electronic concept, a process that involves the loss of electrons by a substance is called oxidation. The substance is said to be oxidized. e.g Li → Li+1 + e–
According to the modern electronic concept, a process that involves the gain of electrons by a substance Is called reduction. The substance is being reduced. e.g N + 3e– → N-3
Q.3 a) What is corrosion? Explain rusting of iron as an example of corrosion.
Answer: Corrosion “Corrosion is slow and continuous eating away of a metal by the environment or surrounding.“ It is an oxidation-reduction process which takes place by the action of air in the presence of moisture with the metals. The terms corrosion and rust are synonymous. The word rust is used more specifically for “iron.” Rusting of iron The corrosion of iron is commonly known as rusting. condition for rusting: The important condition for rusting is moisture and air. There will be no rusting in water vapour free of air or air free of water vapours. Iron rusts by combining with oxygen in the presence of water to form brown hydrated mass ferric oxide (Fe2O3. H2O). Anode: Stains and dents on the surface of iron provide the sites for this process to occur. This region is called anodic region and following oxidation reaction takes place here: 2Fe → 2Fe2+ + 4e– This loss of electrons damages the object. The free electrons move through iron sheet until they reach to a region of relatively high O2 concentration near the surface surrounded by water layer. Cathode: A region of relatively high O2 concentration near the surface surrounded by water layer acts as cathode and electrons reduce oxygen molecule. 2H2O +O2 + 4e– → 4OH–
Fe2+is further oxidized by atmospheric oxygen to form hydrated Fe(III) oxide (rust) Fe2+→ Fe3+ + e– Fe3 + 3OH– → Fe(OH)3 Fe(OH)3 produced iscalled rust. The rust mass produce is soft and porous in nature and therefore cannot prevent further atmospheric action.
Q.3 b) Differentiate between electrolytic cell and voltaic cell.
An electrolytic cell converts electrical energy into chemical energy.
A voltaic cell converts chemical energy into electrical energy.
The redox reaction is non-spontaneous and electrical energy has to be supplied to initiate the reaction.
The redox reaction is spontaneous and is responsible for the production of electrical energy.
Both the electrodes are placed in the same container in the solution of molten electrolyte. Salt bridge is not required.
The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. Salt bridge is required.
The anode is positive and the cathode is the negative electrode.
The anode is negative and the cathode is the positive electrode.
The reaction at the anode is oxidation and that at the cathode is reduction.
The reaction at the anode is oxidation and that at the cathode is reduction.
The external battery supplies the electrons. They enter through the cathode and come out through the anode.
The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit.
Q.3 c) Discuss the method of recovering/extracting of metal from its ore.
Answer: Electrolytic refining of copper Pure copper is very good conductor of electricity and is used in electrical instrument. Copper is purified by electrorefining. Large blocks of blistered copper (Impure) 99% pure acts as the anode and a pure copper plate acts as a cathode. Copper sulphate (CuSO4) and Sulphuric acid (H2SO4) solution is used as an electrolyte. The operation is performed at 50°C and applied voltage of about 0.3 volts and optimum current density used is 160-400 A/m2. Reaction at Anode Oxidation reaction takes place at the anode. Copper atoms from the impure copper lose electrons to the anode and dissolve in solution as copper ions: Cu → Cu2+ + 2e– As the process of electrolysis is continued, copper from anode goes into solution. Traces of more active metals like Zn, Fe etc are also dissolved. The less active metals e.g Au, Ag remains un-dissolved and settled at the bottom of cell as “Anode sludge/ Anode mud“, which is further processed to recover these precious metals. The voltage and temperature condition are such that only copper is deposited at cathode. Reaction at Cathode Reduction reactions take place at the cathode. The copper ions present in the solution are attracted to the cathode. Where they gain electrons from the cathode and become neutral and deposit on the cathode. Cu2+ + 2e– → Cu In the process, impure copper is eaten away and purified copper atoms deposit on the cathode. By electrolytic refining 99.99% pure copper is obtained.
Q.4) Discuss the preparation of Sodium Hydroxide (NaOH) from the Brine along with diagram and reactions at cathode and anode.
Answer: Manufacture of NaOH NaOH (caustic soda) is produced on a commercial scale by the electrolysis of an aqueous concentrated solution of NaCl called brine. For this purpose, a special type of cell called nelson’s cell is used.
Construction of cell The cell consists of a U-shaped perforated iron vessel, which acts as a cathode. The cathode is coated inside with a diaphragm of asbestos. A carbon rod is suspended inside the U-shaped vessel which acts as an anode. Asbestos separates the anode from the cathode.
Working of cell During the electrolysis, the chlorine is produced at the anode. It is collected at the chlorine outlet. Hydrogen gas is produced at the cathode. It is collected at the hydrogen outlet. During this reaction, sodium hydroxide is also produced. The sodium hydroxide is collected in the catch basin, placed under the U-shaped tube. In this process, the hydrogen, chlorine and sodium hydroxide is produced at the same time. The saturated brine solution ionizes as follows: 2NaCl → 2Na+ + 2Cl– 2H2O ⇋ H+ + OH– When the electrodes are connected to the battery, the positive ions, Na+ and H+ move towards the cathode. Since H+ have a great tendency to pick up electrons from the cathode as compared to Na+, therefore, H+ ions pick up electrons to form H2 gas. Na+ ions are not reduced instead they combine with OH– ions, present in the solution to form caustic soda (NaOH) which make the solution alkaline, while Cl– ions move towards the anode where they give electrons to the electrode. At anode (oxidation) 2Cl– → Cl2 + 2e– At cathode (reduction) 2H+ + 2e– → 2H 2Na+ + 2OH– → 2NaOH Overall reaction 2NaCl + 2H2O → 2NaOH + H2 + Cl2