In principle, O can have a minimum O.N. Answer: Writing the O.N. of C2 = 3 (+1) + x + 1 (-1) = 0 or x = -2 C2 is, however, attached to one OH (O.N. Write a balanced ionic equation for the reaction. of C in cyanogen, (CN)2 = 2 (x – 3) = 0 or x = +3 O.N. (iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion). What is meant by electrochemical series? (iv) In aqueous solution, CuCl2 ionises as follows: 160 g of 02 produce NO = 120 g (c) 4BCl3(g) +3LiAlH4(s) ——> 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) Thus, the O.N. Answer: At cathode there is gain of electrons. Since the electrode potentials increase in the oder; K+/K (-2.93 V), Mg2+/Mg (-2.37 V), Cr3+/Cr (-0.74 V), Hg2+/Hg (0.79 V), Ag+/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag. Answer: EMF of a cell is the difference in the electrode potentials of the two electrodes in a cell when no current flows through the cell. Balance the following equation in basic medium by ion electron method and oxidation number method and identify the oxidising agent and the reducing agent. Hydrogen electrode can be made. At anode there is loss of electrons. (a) HNO3 acts only as an oxidising agent while HNO3 can act both as reducing and oxidising agent. Question 12. Thus, F2 is the best oxidant. In order to do this, the half-reaction method can be used. H2O2(aq) +2Fe2+(aq) +2H+(aq) ——-> 2Fe3+(aq) + 2H2O(l) Question 4. (a) HCHO (b) CH2Cl2 (c)C12H22O21 (d) C6H12O of Cu decreases from + 2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H20. 4. Since the oxidation potential of SO4 is expected to be much lower (since it involved cleavage of many bonds as compared to those in H20) than that of HjO molecules, therefore, at the anode, it is H2O molecules (rather than SO42- ions) which are oxidised to evolve O2 gas. To do so, Eq. If, however, excess of 02 is used, Na2O2 is formed in which the oxidation state of O is -1 which is higher than -2. Question 28. F2(g) + 2Cr(aq) ———–> 2F–(aq) + Cl2(g); F2(g) + 2Br–(aq) ———-> 2F–(aq) + Br2 (Z) However, if formed, the compound acts as a very strong oxidising agent. For example, number method. Answer: In a galvanic cell due to redox reaction released energy gets converted into the electrical energy. Question 9. Click hereto get an answer to your question ️ Balance the following equations by the ion electron method:a. MnO4^ + Cl^ + H^⊕ Mn^2 + + H2O + Cl2 b. Cr2O7^2 - + I^ + H^⊕ Cr^3 + + H2O + I2 c. H^⊕ + SO4^2 - + I^ H2S + H2O + I2 d. MnO4^ + Fe^2 + Mn^2 + + Fe^3 + + H2O (d) Cr2O72- (aq) + S02 (g)——> Cr3+ (aq) + SO42-(aq) (in acidic solution) Write the oxidation number of Cr above its symbol and that of H2O above its formula. CU2++ 2e– ———> Cu(s); E° = +0.34 V (a) an oxidizing agent (b) a reducing agent Question 20. 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