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Byju's Answer

Standard VII

Chemistry

Faraday's First Law

The amount of...

Question

The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO $_{3}$ for 1 hour is closest to :

2 g

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5 g

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108 g

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11 g

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Solution

The correct option is A
2 g

$W = \frac{E}{96500} \times I \times t$
$W = \frac{108}{96500} \times 0.5 \times 3600 = 2 g m$

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Similar questions

The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO $_{3}$ for 1 hour is closest to :

Q. The amount of

A g

(atomic mass

= 108

) deposited at the cathode when a current of

0.5 a m p

is passed through a solution of

A g N O_{3}

for

1

hour is closest to

Q. Calculate the mass of

A g

deposited at cathode when a current of 2 amperes was passed through a solution of

A g N O_{3}

for 15 minutes.

Q. Calculate the mass of Ag deposited at cathode when a current of 2A was passed through a solution of

A g {N O}_{3}

for 15 min.
(Given : Molar mass of

A g = 108 g {m o l}^{- 1}

1 F = 96500 C {m o l}^{- 1}

Q. A current of

0.5

ampere when passed through

A g {N O}_{3}

solution for

193

seconds deposited

0.108 g

A g

. The equivalent weight of

A g

is:

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The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO3 for 1 hour is closest to :

The correct option is A 2 g W=E96500×I×t W=10896500×0.5×3600=2gm

The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO $_{3}$ for 1 hour is closest to :

The correct option is A
2 g

$W = \frac{E}{96500} \times I \times t$
$W = \frac{108}{96500} \times 0.5 \times 3600 = 2 g m$