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Mass deposited, current or time in electrolysis (Faraday's first law)

This type of question, asked 4 times: 25 Jun 2022, 13 Apr 2023, 29 Jan 2024, 22 Jan 2025. All 5 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    145
    A solution of Fe2(SO4)3Fe_{2}\left( SO_{4} \right)_{3} is electrolyzed for ' xx ' min with a current of 1.5 A1.5\text{ }A to deposit 0.3482 g0.3482\text{ }g of FeFe. The value of xx is [nearest integer] Given: 1 F=96500Cmol−11\text{ }F = 96500Cmol^{- 1} Atomic mass of Fe=56 g mol−1Fe = 56\text{ }g{\text{ }mol}^{- 1}

  2. Question 2

    13 Apr 2023

    60
    A metal surface of 100 cm2100{\text{ }cm}^{2} area has to be coated with nickel layer of thickness .001 mm.001\text{ }mm. A current of 2 A2\text{ }A was passed through a solution of Ni(NO3)2Ni\left( NO_{3} \right)_{2} for ' xx ' seconds to coat the desired layer. The value of xx is _______(Nearest integer) ( ρNi\rho_{Ni} (density of Nickel) is 10gmL−110gmL^{- 1}, Molar mass of Nickel is 60gmol−1 F=96500Cmol−160gmol^{- 1}\text{ }F = 96500Cmol^{- 1} )

  3. Question 3

    29 Jan 2024

    149
    A constant current was passed through a solution of AuCl4−AuCl_{4}^{-}ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g1.314\text{ }g. The total charge passed through the solution is _______ ×10−2 F\times10^{- 2}\text{ }F. (Given atomic mass of Au=197Au = 197)

  4. Question 4

    29 Jan 2024

    52
    The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A0.015\text{ }A for 15 minutes is _____ ×10−4 g\times10^{- 4}\text{ }g. (Atomic mass of zinc =65.4amu= 65.4amu )

  5. Question 5

    22 Jan 2025

    26
    A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is ___\_\_\_ .[Given : molar mass of aluminium and chlorine are 27 g mol−127\text{ }g{\text{ }mol}^{- 1} and 35.5 g mol−135.5\text{ }g{\text{ }mol}^{- 1} respectively, Faraday constant  =96500Cmol−1]\left. \ = 96500Cmol^{- 1} \right\rbrack.

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