Question - 21 : -
Consider the elements : Cs, Ne, I, F
1.   Identify the elementthat exhibits only -ve oxidation state.
2.   Identify the elementthat exhibits only +ve oxidation state.
3.   Identify the elementthat exhibits +ve and -ve oxidation states.
4.   Identify the elementthat neither exhibits + ve and -ve oxidation states.

Answer - 21 : -

1. F (fluorine) exhibitsonly -ve oxidation state (- 1) in its compounds because it is the mostelectronegative element.

2. Cs (cesium) exhibitsonly +ve oxidation state (+ 1) its compounds because it is the mostelectropositive element.

3. Iodine has sevenelectrons in the valance shell as well as vacant 5d orbital to which electronsfrom 5p and 5s orbitals can be shifted. Therefore, the elements exhibits – 1oxidation state as well as variable positive oxidation states of + 1, +3, +5and +7 in its compounds.

4. Ne (neon) neitherexhibits +ve nor -ve oxidation states because it is a noble gas element withcompletely filled orbitals (Is² 2s²2p⁶).Moreover, it has no vacant d-orbitals.

Question - 22 : - Chlorine is used to purify drinking water. Excess of chlorine isharmful. The excess of chlorine is removed by treating with sulphur dioxide.Present a balanced equation for this redox change taking place.
(P.I.S.A. Based)

Answer - 22 : - Chlorine reacts with sulphur dioxide in the presence of water as follows
Cl₂ (aq) + SO₂ (aq) + H₂O (l) → Cl^– (aq)+ SO₄^2- (aq)

Question - 23 : - From the periodic table, select three non-metals and three metals whichcan show disproportionation reactions.

Answer - 23 : - Non-metals and metals which show variable oxidation states can take partin the disproportionation reactions. For examples,
Non-metals : The non-metalsshowing disproportionation reactions states are : P₄, Cl₂ andS₈.

Question - 24 : - In the Ostwald process for the manufacture of nitric acid, the firststep involves the oxidation of ammonia gas by oxygen gas to give nitric oxideand steam. What is the maximum weight of nitric oxide that can be obtainedstarting only with 10.0 g of ammonia and reacting with 20.0 g of oxygen ?

Answer - 24 : -

Question - 25 : - Using the standard electrode potentials given in the table, predict ifthe 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 Fc³⁺(aq)
(f) Br₂(aq) and Fe²⁺(aq).

Answer - 25 : - A particular reaction can be feasible if e.m.f. of the cell based on theE° values is positive. Keeping this in mind, let us predict the feasibility ofthe reactions.

Question - 26 : - Arrange the following metals in the order in which they displace eachother from their salts. Al, Cu, Fe, Mg and Zn

Answer - 26 : - This is based upon the relative positions of these metals in theactivity series. The metal placed lower in the series can displace the metalsoccupying a higher position present as its salt. Based upon this, the correctorder is : Mg, Al, Zn, Fe, Cu.

Question - 27 : - Given the standard electrode potentials
K+/K = – 2.93 V, Ag+/Ag = 0.80V
Hg²+/Hg = 0.79 V ; Mg²⁺/Mg = – 2.37 V, Cr³⁺/Cr= – 0.74 Y
Arrange these metals in increasing order of their reducing power.

Answer - 27 : -

If may be noted thatlesser the E° value for an electrode, more will be reducing power. Theincreasing order of reducing power is :
Ag < Hg < Cr < Mg < K.

Question - 28 : - Depict the galvanic cell in which the reaction Zn(s) + 2Ag ⁺(aq)→Zn²⁺(aq) + 2Ag(s) takes place. Further show :
(1) which electrode is negatively charged.
(2) the carriers of the current in the cell.
(3) individual reaction at each electrode.

Answer - 28 : - The galvanic cell for the reaction is

· Zinc electrode (anode)is negatively charged.

· Current flows fromsilver to zinc in outer circuit

· Anode : Zn (s) → Zn²⁺(aq)+ 2e^–(oxidation)
Cathode : 2Ag⁺ (ag)+ 2e^– → 2Ag(s) (reduction)

Free - Previous Years Question Papers

Chapter 8 Redox Reactions Solutions

Subscribe Now!