What is correct for λ m NH 4 OH 0 ?

λ m NH 4 Cl 0 + λ m NaOH 0 - λ m NaCl 0

λ m NH 4 Cl 0 + λ m NaCl 0 - λ m NaOH 0

λ m NaOH 0 + λ m NH 4 Cl 0 - λ m HCl 0

λ m NaCl 0 + λ m NH 4 Cl 0 - λ m NaOH 0

On charging of lead storage cell

PbO 2 is deposited on one of the electrode

H 2 SO 4 of solution is consumed

On which of the following factor ionic conductivity of solution does not depend?

Size of molecules produce in solution

Which of the following will be possible in electrochemical cell obtained from E Cl 2 2 Cl - 0 = 1 . 36 V and E Br 2 2 Br - 0 = 1 . 09 V ?

2 Br - + Cl 2 → Br 2 + 2 Cl -

2 Cl - + 2 Br - → Cl 2 + Br 2

Br 2 + 2 Cl - → 2 Br - + Cl 2

Cl 2 + Br 2 → 2 Cl - + 2 Br -

How many spoons can be electroplated by silver when 7 ampere current is passed through electrolyte cell of AgNO 3 for 1 . 93 hours? 0 . 01 gm Ag layer is deposited on each spoon. ( Ag = 108 g . mol - 1 )

Given: I = 7 A t = 1 . 93 h = 1 . 93 × 3600 s The total charge passed through the electrolyte: Q = It = 7 × 1 . 93 × 3600 C = 48636 C The reduction process taking place: Ag + + e - → Ag Now, 1 mol of electron charge, i.e. 96500 C will produce 108 g Ag The moles of electrons passed during electroplating = 48636 96500 = 0 . 504 mol Therefore, total silver deposited = 0 . 504 × 108 = 54 . 432 g Coating of Ag required on one spoon = 0 . 01 g The number of spoons coated = Mass of silver deposited Coating on 1 spoon = 54 . 432 0 . 01 = 5443 Hence, a total of 5443 spoons can be coated.

The potential of the given following cell is 0 . 092 volt at 298 K temperature. Calculate the pH of HCl solution ( E Sn Sn 2 + ∘ = + 0 . 14 volt). Sn | Sn 2 + ( 0 . 05 M ) | | H + ( xM ) | H 2 ( g ) ( 1 bar ) | Pt

Given: E cell = 0 . 092 V at 298 K Anode: Sn → Sn 2 + + 2 e - E Sn 2 + Sn 0 = − 0 . 14 V Cathode: 2 H + + 2 e - → H 2 E H + H 2 0 = 0 V The complete cell reaction is: Sn + 2 H + → Sn 2 + + H 2 ∴ E cell 0 = E H + H 2 0 − E Sn 2 + Sn 0 = 0 − ( − 0 . 14 ) = 0 . 14 V Using Nernst equation for the cell reaction: E cell = E cell 0 − RT nF ln [ Sn 2 + ] [ H + ] 2 0 . 092 = 0 . 14 − 0 . 059 2 ln 0 . 05 x 2 ⇒ x = 0 . 1 Calculation of pH : pH = − log ( [ H + ] ) = − log ( 0 . 1 ) = 1

Three electrolytic cells A , B and C containing solution NiSO 4 , AgNO 3 and CuSO 4 respectively are connected in series. A steady current of 1 . 5 amperes was passed through them until 1 . 45 g of silver deposited at the cathode of cell B . How long did the current flow? What mass of copper and nickel were deposited? Atomic mass of Ag = 108 u , Ni = 58 . 7 u , Cu = 63 . 5 u .

For electrolyte B : A g + a q + e - → A g s 108 g i.e., 1 m o l e of A g is deposited by 1 F charge i.e., 108 g of A g is deposited by 96500 C . Therefore, 1.45 g of A g is deposited by = 96500 × 1.45 108 = 1295 .6 C Given, Current = 1.5 A , Therefore time, t = 1295.6 1.5 s = 863.7 s = 864 60 min ⁡ = 14.40 m i n . For electrolyte A : Ni 2 + a q + 2 e - → Ni s 58 . 7 g 2 × 96500 C of charge deposits 58 . 7 g of Ni Therefore, 1295. 6 C of charge will deposit 58 . 7 × 1295. 6 2 × 96500 g = 0.394 g of Ni For electrolyte C : C u 2 + a q + 2 e - → Cu s 63.5 g ⁡ 1 m o l e of C u is deposited by 2 F charge. i.e., 2 × 96500 C of charge deposits 63.5 g of C u Therefore, 1295. 6 C charge will deposit 63.5 × 1295. 6 2 × 96500 g = 0.426 g of C u .

Resistance of conductivity cell filled with 0 . 1 mol L - 1 KCl solution is 100 ohm . If the resistance of the same cell when filled with 0 . 03 mol L - 1 KCl solution is 520 ohm , calculate the conductivity and molar conductivity of 0 . 03 mol L - 1 KCl solution. The conductivity of 0 . 1 mol L - 1 KCl solution is 1 . 29 S m - 1 .

Given, R 1 = 100 ohm , R 2 = 520 ohm , Κ 1 = 1 . 29 Sm - 1 , C 1 = 0 . 1 mol L - 1 , C 2 = 0 . 03 mol L - 1 and Κ 2 = ? We know that, Cell constant G * = Conductivity Κ × Resistance R G * = Κ 1 × R 1 = 1 . 29 × 100 = 129 m - 1 Similarly, G * = Κ 2 × R 2 , Κ 2 = G * R 2 = 129 520 = 0 . 248 Sm - 1 We know that, Molar conductivity ∧ m 2 = 1000 Κ 2 C 2 = 0 . 248 × 1000 0 . 03 = 8 . 27 × 10 3 S m 2 mol - 1 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Electrochemistry>Gujarat Board-2020>Q 12

EASY

12th Gujarat Board

IMPORTANT

Earn 100

Which is increasing order of reducing power of the following metals on the basis of standard electrode potential?

${E^{\circ}}_{{Ag}^{+} / Ag} = 0.80 V, {E^{\circ}}_{{Mg}^{2 +} / Mg} = - 2.37 V, {E^{\circ}}_{{Hg}^{2 +} / Hg} = 0.79 V, {E^{\circ}}_{{Cr}^{3 +} / Cr} = - 0.74 V$

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