CBSE · NCERT · Class 12 Chemistry · Chapter 2

NCERT Solutions: Class 12 Chemistry Chapter 2 - Electrochemistry

14 textbook Q&A14 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Electrochemistry, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercises 14
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1Exercises14 questions
Q.2.1How would you determine the standard electrode potential of the system Mg2+|Mg?v
Solution

Set up Mg(s)|Mg2+(1 M)||H+(1 M)|H2(1 bar), Pt. Magnesium acts as the anode and the standard hydrogen electrode acts as the cathode. Since E°SHE = 0, E°cell = 0 - E°Mg2+/Mg. Thus E°Mg2+/Mg = -E°cell measured under standard conditions.

Answer:

Connect Mg2+|Mg as one half-cell with the standard hydrogen electrode and measure the standard cell emf.

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Q.2.2Can you store copper sulphate solutions in a zinc pot?v
Solution

Zinc is more reactive than copper and has a lower standard reduction potential. It will displace copper from copper sulphate solution: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Therefore copper sulphate solution cannot be stored in a zinc pot.

Answer:

No.

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Q.2.3Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.v
Solution

Fe3+/Fe2+ has E° = +0.77 V. Any oxidising agent with a higher reduction potential can oxidise Fe2+ to Fe3+. Suitable examples include MnO4-/Mn2+ in acid, Cr2O7^2-/Cr3+ in acid, and Cl2/Cl-.

Answer:

Examples are acidified KMnO4, acidified K2Cr2O7 and chlorine.

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Q.2.4Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.v
Solution

For 2H+(aq) + 2e- → H2(g), with pH2 = 1 bar, E = E° - (0.0591/2)log(1/[H+]^2) = -0.0591 pH. At pH = 10, E = -0.0591 x 10 = -0.591 V.

Answer:

-0.591 V.

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Q.2.5Calculate the emf of the cell in which the following reaction takes place: Mg(s) + Cu2+(0.0001 M) → Mg2+(0.001 M) + Cu(s) Given that E°cell = 2.71 V.v
Solution

For Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s), n = 2 and Q = [Mg2+]/[Cu2+] = 0.001/0.0001 = 10. Ecell = E°cell - (0.0591/2)log Q = 2.71 - (0.0591/2)log10 = 2.68 V.

Answer:

2.68 V.

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Q.2.6The cell in which the following reaction occurs: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.v
Solution

Here n = 2. ΔrG° = -nFE°cell = -2 x 96485 x 0.236 = -4.55 x 10^4 J mol^-1 = -45.5 kJ mol^-1. Also E°cell = (0.0591/n)log K at 298 K, so log K = nE°/0.0591 = 2(0.236)/0.0591 = 7.99. Hence K ≈ 9.7 x 10^7.

Answer:

ΔrG° = -45.5 kJ mol^-1; K = 9.7 x 10^7.

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Q.2.7Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.v
Solution

Conductivity κ measures how well a solution conducts electricity and has units S m^-1 or S cm^-1. Molar conductivity Λm = κ x 1000/C when κ is in S cm^-1 and C in mol L^-1. Conductivity decreases on dilution because ions per unit volume decrease. Molar conductivity increases on dilution because interionic interactions decrease and ions move more freely; for weak electrolytes it increases sharply due to increased dissociation.

Answer:

Conductivity is conductance of a solution of unit length and unit cross-section. Molar conductivity is conductivity of the volume of solution containing one mole of electrolyte.

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Q.2.8The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.v
Solution

Λm = κ x 1000/C = 0.0248 x 1000/0.20 = 124 S cm^2 mol^-1.

Answer:

124 S cm^2 mol^-1.

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Q.2.11The conductivity of 0.00241 M acetic acid is 7.896 x 10^-5 S cm^-1. Calculate its molar conductivity. If Λm° for acetic acid is 390.5 S cm^2 mol^-1, what is its dissociation constant?v
Solution

Λm = κ x 1000/C = (7.896 x 10^-5 x 1000)/0.00241 = 32.76 S cm^2 mol^-1. Degree of dissociation α = Λm/Λm° = 32.76/390.5 = 0.0839. Ka = Cα^2/(1 - α) = 0.00241(0.0839)^2/(0.9161) = 1.85 x 10^-5.

Answer:

Λm = 32.76 S cm^2 mol^-1; Ka = 1.85 x 10^-5.

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Q.2.12How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al (ii) 1 mol of Cu2+ to Cu (iii) 1 mol of MnO4- to Mn2+?v
Solution

Al3+ + 3e- → Al requires 3 mol electrons per mole Al3+, so charge = 3F. Cu2+ + 2e- → Cu requires 2F. In acidic solution, MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, so 5F is required.

Answer:

(i) 3 F; (ii) 2 F; (iii) 5 F.

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Q.2.13How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2 (ii) 40.0 g of Al from molten Al2O3?v
Solution

For Ca2+ + 2e- → Ca, 1 mol Ca requires 2F. 20.0 g Ca = 0.500 mol, so charge = 0.500 x 2 = 1.0 F. For Al3+ + 3e- → Al, 40.0 g Al = 40.0/27.0 = 1.48 mol, so charge = 1.48 x 3 = 4.44 F.

Answer:

(i) 1.0 F; (ii) 4.44 F.

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Q.2.14How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2 (ii) 1 mol of FeO to Fe2O3?v
Solution

For water oxidation, 2H2O → O2 + 4H+ + 4e-. Therefore 1 mol H2O corresponds to 2 mol electrons = 2F = 1.93 x 10^5 C. In FeO to Fe2O3, Fe2+ is oxidised to Fe3+, requiring 1 electron per Fe. Therefore 1 mol FeO requires 1F = 9.65 x 10^4 C.

Answer:

(i) 2 F = 1.93 x 10^5 C; (ii) 1 F = 9.65 x 10^4 C.

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Q.2.15A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?v
Solution

Ni2+ + 2e- → Ni. Charge passed = It = 5 x 20 x 60 = 6000 C. Mass deposited = (MIt)/(nF) = (58.7 x 6000)/(2 x 96500) = 1.82 g.

Answer:

1.82 g.

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Q.2.16Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, 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 zinc were deposited?v
Solution

For Ag+ + e- → Ag, charge required for 1.45 g Ag is Q = (1.45/107.9)F = 1297 C. Time = Q/I = 1297/1.5 = 865 s. The same charge passes through all cells. Cu2+ + 2e- → Cu gives mass Cu = (63.5/2F)Q = 0.427 g. Zn2+ + 2e- → Zn gives mass Zn = (65.4/2F)Q = 0.439 g.

Answer:

Time = 865 s; Cu deposited = 0.427 g; Zn deposited = 0.439 g.

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