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Electrochemistry

Short Answer Type

131. Write the cell reactions which occur in lead storage battery (i) when the battery is in use and (ii) when the battery is on charging.

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132. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 s cm^–1. Calculate its molar conductivity.

We have given that
Molarity of solution, M = 0.20
conductivity, i.e., specific conducitivity = k = 0.248 s cm^–1= 2.48 x 10^–2 ohm^–1cm^–1

Molar conductivity, λ_{m} = \frac{1000 k}{M} {ohm}^{- 1} {cm}^{2} {mol}^{- 1} = \frac{1000 \times 2.48 \times 10^{- 2}}{0.20} = 124.0 s {cm}^{2} {mol}^{- 1}

Thus, molar conductivity,λ_m = 124.0 s cm²mol^–1

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133.

How much charge is required for the following reductions:
1 mol of Al³⁺ to Al?

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134.

How much charge is required for the following reductions:
1 mol of Cu²⁺ to Cu?

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135.

How much charge is required for the following reductions:
1 mol of MnO₄^– to Mn²⁺?

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136. How much electricity in terms of Faraday is required to produce
20.0 g of Ca from molten CaCl₂?

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137. How much electricity in terms of Faraday is required to produce
40.0 g of Al from molten Al₂O₃?

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138. How much electricity is required in coulomb for the oxidation of
1 mol of H₂O to O₂ ?

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139. How much electricity is required in coulomb for the oxidation of
1 mol of FeO to Fe₂O₃

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140. A 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?

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

132. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 s cm–1. Calculate its molar conductivity.

132. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 s cm^–1. Calculate its molar conductivity.