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) → ClO – 2(aq) + O2(g) + H + (aq)
(a) The O.N. (oxidation number) of P decreases from 0 in P4 to -3 in PH3 and increases from 0 in P4 to + 2 in HPO-2. Hence, P4 acts both as an oxidizing agent as well as a reducing agent in this reaction.
Ion-electron method:
The oxidation half equation is:
P4(s) → H2PO-(aq)
The P atom is balanced as:
P0 4(s) → 4H2P+1O-(aq)
The O.N. is balanced by adding 4 electrons as:
P4(s) → 4H2PO-(aq) + 4e-
The charge is balanced by adding 8OH- as:
P4(s) + 8OH - (aq) → 4H2PO-2(aq)
The O and H atoms are already balanced. The reduction half equation is:
P4(s) → PH3(g)
The P atom is balanced as
P04(s) → 4 P-3H3(g)
The O.N. is balanced by adding 12 electrons as:
P4(s) + 12e- → 4 PH3(g)
The charge is balanced by adding 12OH- as:
P4(s) + 12e- → 4 PH3(g) + 12OH-(aq) .....(i)
The O and H atoms are balanced by adding 12H2O as:
P4(s) + 12H2O(l) + 12e- → 4 PH3(g) + 12OH-(aq) -- (ii)
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
P4(s) + 3OH-(aq) + 3H2O → PH3 + 3H2PO-2(aq)
(b)
The oxidation number of N increases from -2 in N2H4 to +2 in NO and the oxidation number of Cl decreases from + 5 in CIO-3 to -1 in Cl-. Hence, in this reaction, N2H4 is the reducing agent and CIO-3 is the oxidizing agent. Ion-electron method:
The oxidation half equation is:
N-22 H4(l) → N+2 O(g)
The N atoms are balanced as:
N2H4(l) → 2NO(g)
The oxidation number is balanced by adding 8 electrons as:
N2H4(l) → 2NO(g) + 8e-
The charge is balanced by adding 8 OH-ions as:
N2H4(l) + 8OH-(aq) → 2NO(g) + 8e-
The O atoms are balanced by adding 6H2O as:
N2H4(l) + 8OH-(aq) → 2NO(g) + 6H2O(l) + 8e- .... (i)
The reduction half equation is:
C+5IO-3(aq) → C-1l-(aq)
The oxidation number is balanced by adding 6 electrons as:
CIO-3(aq) + 6e- → Cl-(aq)
The charge is balanced by adding 6OH- ions as:
CIO-3(aq) + 6e- → Cl-(aq) + 6OH-(aq)
The O atoms are balanced by adding 3H2O as:
CIO-3(aq) + 3H2O(l) + 6e- → Cl-(aq) + 6OH-(aq) .... (ii)
The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq) + 6H2O(l)
Oxidation number method:
Total decrease in oxidation number of N = 2 × 4 = 8
Total increase in oxidation number of Cl = 1 × 6 = 6
On multiplying N2H4 with 3 and CIO-3 with 4 to balance the increase and decrease in O.N., we get:
3N2H4(l) + 4CIO-3(aq) → NO(g) + Cl-(aq)
The N and Cl atoms are balanced as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq)
The O atoms are balanced by adding 6H2O as:
3N2H4(l) + 4CIO-3(aq) → 6NO(g) + 4Cl-(aq) + 6H2O(l)
This is the required balanced equation.
(c)
The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in CIO-2and the oxidation number of O increases from -1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.
Ion-electron method:
The oxidation half equation is:
H2O-12(aq) → O02(g)
The oxidation number is balanced by adding 2 electrons as:
H2O2(aq) → O2(g) + 2e-
The charge is balanced by adding 2OH-ions as:
H2O2(aq) + 2OH-(aq) → O2(g) + 2e-
The oxygen atoms are balanced by adding 2H2O as:
H2O2(aq) + 2OH-(aq) → O2(g) + 2H2O(l) + 2e- ... (i)
The reduction half equation is:
C+7l2O7(g) → C+3lO-2(g)
The Cl atoms are balanced as:
Cl2O7(g) → 2ClO-2(g)
The oxidation number is balanced by adding 8 electrons as:
Cl2O7(g) + 8e- → 2ClO-2(g)
The charge is balanced by adding 6OH- as:
Cl2O7(g) + 8e- → 2ClO-2(g) + 6OH- (aq)
The oxygen atoms are balanced by adding 3H2O as:
Cl2O7(g) + 3H2O(l) + 8e- → 2ClO-2(g) + 6OH- (aq) .... (ii)
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
Cl2O7(g) + 4H2O2(aq) + 2OH- (aq) → 2ClO-2(aq) + 4O2(g) + 5H2O(l)
Oxidation number method:
Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8
Total increase in oxidation number of H2O2 = 2 × 1 = 2
By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:
Cl2O7(g) + 4H2O2(aq) → CIO-2(aq) + 4O2(g)
The Cl atoms are balanced as:
Cl2O7(g) + 4H2O2(aq) → 2CIO-2(aq) + 4O2(g)
The O atoms are balanced by adding 3H2O as:
Cl2O7(g) + 4H2O2(aq) → 2CIO-2(aq) + 4O2(g) + 3H2O(l)
The H atoms are balanced by adding 2OH- and 2H2O as:
Cl2O7(g) + 4H2O2(aq) + 2OH-(aq) → 2CIO-2(aq) + 4O2(g) + 5H2O(l)
This is the required balanced equation.
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
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) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
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.
In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1
An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Suggest reasons why the B-F bond lengths in BF3 (130 pm) and BF-4(143 pm) differ.
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Why does the following reaction occur ?
XeO4– 6(aq) + 2F – (aq) + 6H + (aq) → XeO3(g) + F2(g) + 3H2O(l)
What conclusion about the compound Na4XeO6 (of which XeO4– 6 is a part) can be drawn from the reaction.
Thanks for our information