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Mass of copper deposited by the passage ...

Mass of copper deposited by the passage of 2 A of current for 965 seconds through a 2M solution of `CuSO_(4)` is (At.mass of Cu = 63.5)

A

`0.325` g

B

0.635 g

C

1 g

D

1.2 g

Text Solution

Verified by Experts

The correct Answer is:
B
W = ? , Z = 2 A , t = 965 sec
`W = Z xx ixx t ` , But `Z = (E)/(96500)`
[E = Eq. mass of `Cu = (63.5)/(2)` in `CuSO_(4)`, Cu is in + 2 state]
`= (63.5)/(2) xx (2 xx 965)/(96500) = 0.635` g
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Knowledge Check

  • Mass of copper deposited by the passage of 2 A of current for 965 s through a 2 M solution of CuSO_(4) is (At. Mass of Cu= 63.5) __________.

    A
    0.325 g
    B
    0.635 g
    C
    1 g
    D
    1.2 g

  • The mass of copper deposited from a solution of CuSO_4 by passage of 45 A current for 965 second is (Mol . Wt. pf Copper = 63.5 ).

    A
    ` 15. 875 g`
    B
    ` 1. 585 g`
    C
    ` 4825 g`
    D
    ` 96 500 g`

  • What weight of copper will be deposited by passing 2 Faraday of electricity through a solution of a cupric salt ,(atomic weight of copper= 63.5)?

    A
    Molecular weight
    B
    Electro chemical equivale'nt
    C
    63.5 g
    D
    127.0 g

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    Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of CuSO_(4) . [Molar mass of Cu = 63.5 g mol^(-1) , 1 F = 96500 C mol^(-1) ]

    What is the mass of copper metal produced at cathode during the passage of 2.03A current through the CuSO_4 solution for 1 hour. Molar mass of Cu = 63.5 g mol^-1

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    Calculate the time required to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO_(4) . ("Molar mass of "Cu=63.5g mol^(-1), 1F =96500 C mol^(-1)).

    The amount of copper deposited by the passage of 1 mole of electrons during the electrolysis of aqueous CuSO_(4) solution is

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