Title DPP 10 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 10 1 (c) Equivalent conductance = 1000 × conductance × cell constant normality So, units are, Ω ―1 cm2 equiv―1 or S cm2 equiv―1 . 2 (b) For strong electrolytes the plot of molar conductance (Λm) vs. √C is linear. (Λm)with C for strong electrolyte. 3 (b) Λ∞ eq(NH4 OH) = Λ∞ eq (NH4Cl) + Λ∞ eq (NaOH) ― Λ∞ eq (NaCl) = (149.74 + 248.1 – 126.4) = 271.44 Ω ―1 cm2 eq―1 4 (c) E ° cell = E ° cathode ― E ° anode E ° Ag+ /Ag ― E ° Cu+ /Cu = - 0.80 – 0.34 = + 0.46 5 (c) Ecell = EOPNi/Ni2+ 2 + ERPAu3+/Au = E ° OPNi ― 0.059 2 log [Ni2+] + E ° RPAu + 0.059 3 log [Au 3+] = 0.25 ⎯ 0.059 2 log(1.0) +1.50 + 0.059 3 log 1.0 = 1.75 V 6 (d) The metal placed below in electrochemical series does not react with that metal salt solution which metal is placed above in series. 8 (a) Cell representation is done as follows Anode | Anodic electrolyte || cathodic electrolyte | cathode Strong electrolyte m C Variation of molar conductance Topic :- Electro Chemistry Solutions
(i) Oxidation is loss of electron and it takes place at anode. Reduction is gain of electron and it takes place at cathode. ∴ For cell reaction, Zn + Cu2+ ⟶ Zn2+ + Cu Zn is anode and Cu is cathode. ∴ Cell representation is Zn | Zn2+ || Cu2+ | Cu 9 (c) 2 × 96500 C electricity is used to liberate = 22400 mL O2 at STP ∴ 9.65 × 1000 C electricity will liberate = 22400 × 9.65 × 1000 2 × 96500 = 1120mL 10 (d) Oxidation half-cell ClO― 3 →ClO― 2 + 2e ―; E°cell = ―0.36 V Reduction half-cell ClO― 3 + 2e ―→ClO― 2 ; E°cell = 0.33 V E°cell = 0.33 ― 0.36 = ―0.03 = RT 2F ln K or ― 0.03 = 0.059 2 logK or K = 0.1 2ClO ― 3 ⇌ClO ― 4 + ClO ― 2 0.1 ― 2x x x x 2 (0.1 ― 2x) 2 = 0.1 or x = 1.9 × 10―2 11 (a) Cu is placed above Ag in electrochemical series, hence it can replace Ag from its salts solution. Therefore, the reaction occur as follows
O +1 Cu + AgNO3 Oxidation CuNO3 + Ag 12 (a) E° does not depend on stoichiometry of change. 13 (d) HCl is strong electrolyte and H + has highest conducting power due to Grothus conductance. 14 (d) Fe 2+ + 2e ⟶ Fe; ⎯ ∆G1 = 2 × ( ―0.44) × F Fe 3+ ― + 3e ― ⟶Fe ― ; ― ∆G2 + = 3 × ― ( ― 0.036) × F Fe 2+⟶Fe 3+ + e; ∆G3 = ― 0.88 + 0.108 = 0.772 or 0.772 = 1 × E° × F ∴ Fe 3+ + e ⟶Fe 2+;E° = + 0.772 V Above procedure should be used only when two half reactions on algebraic sum give a third half reaction. 15 (b) Chromium is more electropositive metal than iron. In stainless steel, chromium forms an oxide layer and thus it protects steel from corrosion. 16 (b) Cu 2+ + 2KI ⟶CuI2 + 2K +; 2CuI2 ⟶Cu2I2 + I2 17 (d) Salt bridge is used to remove or eliminate liquid junction potential arised due to different relative speed of ions of electrolytes at the junction of two electrolytes in an electrochemical cell. Thus, a salt bridge such as KCl is placed in between two electrolytes. A salt used for this purpose should have almost same speeds of its cation and anion. 18 (b) wAg = EAg × Q 96500 = 108 × 9.65 96500 = 1.08 × 10―2 g = 10.8 mg 19 (d) Na2S4O6 is The two S atoms which are linked to each other have 0 oxidation number. The oxidation number of other two S-atoms can be calculated as 2x + 2 × 0 + 6 × ―2 = ―2 2x = 12 ― 2 = 10 x = +5 Na + -O S O O S S S O O O -Na +
20 (a) Higher the negative value ofE ° , more is the reducing power. The order of E ° values (negative value) is ―2.37 > ― 0.76 > ― 0.44 (Mg) (Zn) (Fe) ∴ Mg can reduce both Zn2+ andFe2+ . Zn can reduce Fe2+ , but not Mg2+ . Fe cannot reduce Mg and Zn but can oxidize them. ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. C B B C C D A A C D Q. 11 12 13 14 15 16 17 18 19 20 A. A A D D B B D B D A