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Electrochemistry

Short Answer Type

251. Define conductivity and molar conductivity for the solution of an electrolyte.

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252.

On the basis of the standard electrode potential values stated for acid solution. Predict whether Ti⁴⁺ species may be used to oxidise Fe^II to Fe^III.
Reaction: E°/V Ti^IV + e^– → Ti³⁺ + 0.01
Fe³⁺ + e^– → Fe²⁺ + 0.77

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253. What are fuel cells? Write the electrode reactions of a fuel cell which uses the reaction of hydrogen with oxygen.

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254.

Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum.
(i) An aqueous solution of AgNO_3.(ii) An aqueous solution of H₂SO₄.

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Long Answer Type

255. Define the following terms: (i) Cathodic protection, (ii) Electrochemical series, (iii) Cell constant, (iv) Equivalent conductivity, (v) Strong and weak electrolytes.

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256. Explain the working of galvanic cell. How does the electrochemical cell differ from electrolytic cell?

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Short Answer Type

257. What is normal hydrogen electrode? Discuss its uses.

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Long Answer Type

258. What is corrosion? What are the factors which affect corrosion?

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259. Distinguish between: (a) Electrolytes and non-electrolytes, (b) Reduction potential and oxidation potential (c) Primary cells and secondary cells, (d) Specific conductivity and molar conductivity.

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260.
The conductivity of 0.01 M solution of acetic acid at 25°C is 1.63 x 10^–4 s cm^–1. Given:
Λ°_m (HCl) = 426 s cm² mol^–1, Δ°_m (Na AC) = 91.5 cm² mol^–1Λ°_m (NaCl) = 126.5 cm² mol^–1Calculate:
(a) the molar conductivity of acetic acid
(b) the degree of dissociation of acetic acid.
(c) the dissociation constant.
(d) the pH of 0.01 M solution of acetic acid.

a) The molar conductivity of acetic acid given by;

$Λ_{m} = \frac{10^{3} k}{M} = \frac{10^{3} x 1.63 x 10^{- 4}}{0.01} = 16.3 S {cm}^{2} {mol}^{- 1} b) The degree of dissociation (α) is α = \frac{Λ_{m}}{Λ_{m}^{0}} Λ_{m}^{0} = Λ_{m}^{0} (HCl) + Λ_{m}^{0} (NaAc) - Λ_{m}^{0} (NaCl) = 426 + 91 - 126 = 391 S {cm}^{2} {Mol}^{- 1} Therefore α = \frac{Λ_{m}}{Λ_{m}^{0}} = \frac{16.3}{391} = 0.042 c) K = \frac{{Cα}^{2}}{1 - α} = \frac{0.01 x (0.042)^{2}}{1 - 0.042} = 1.84 x 10^{- 5} d) [H^{+}] = C α = 0.01 x 0.042 = 4.2 x 10^{- 4} M p H = - \log [H^{+}] = - \log (4.2 x 10^{- 4}) = 3.38$

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