You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Faraday's Second Law

10800 C of el...

Question

$10800 C$ of electricity through the electrolyte deposited $2.977 g$ of metal with atomic mass $106.4 g m o l^{- 1}$ . The charge on the metal cation is

+ 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $+ 4$
$M^{n +} + n e^{-} \to M$
Mole of metal deposited $= \frac{2.977}{106.4} = 0.27 m o l$
1 mol of M deposited = n Faraday
= n $\times$ 96500 C.
$0.027 m o l = n \times 96500 \times 0.027$
$∴ n \times 96500 \times 0.027 = 10800 C$
$∴ n = 4$

Suggest Corrections

Similar questions

10800 C

of electricity through the electrolyte deposited

2.977 g

of metal with atmic mass

106.4 g m o l^{- 1}

. The charge on the metal cation is:

10800 C

of electricity on passing through the electrolyte solution deposited

2.977

of metal with atomic mass

{106.4 g m o l}^{- 1}

the charge on the metal cation is:

10800C ofelectricity through the electrolyte deposited 2. 977 gram of metal with atomic mass of 106.4 gram per mole find the charge on the metal cations

Q. Calculate the amount of zinc deposited at the cathode when 3 amperes current is passed through the electrolyte for 2 minutes. Also, calculate the equivalent weight and oxidation state of the other metal if 0.245 grams of the metal is deposited by the same quantity of the electricity.

(Atomic weight of metal = 197 g/mole)

Q. A solution of metal salt was electrolyzed with a current of

0.1

A for

160

minutes.

N i

deposited at cathode was found

0.2950

What is the charge on metal ion? (Atomic mass of metal is

58.71

)

Join BYJU'S Learning Program

10800 C of electricity through the electrolyte deposited 2.977 g of metal with atomic mass 106.4 g mol−1. The charge on the metal cation is

The correct option is C +4Mn++ne−→MMole of metal deposited =2.977106.4=0.27mol1 mol of M deposited = n Faraday = n × 96500 C.0.027mol=n×96500×0.027∴n×96500×0.027=10800C∴n=4

$10800 C$ of electricity through the electrolyte deposited $2.977 g$ of metal with atomic mass $106.4 g m o l^{- 1}$ . The charge on the metal cation is

The correct option is C $+ 4$
$M^{n +} + n e^{-} \to M$
Mole of metal deposited $= \frac{2.977}{106.4} = 0.27 m o l$
1 mol of M deposited = n Faraday
= n $\times$ 96500 C.
$0.027 m o l = n \times 96500 \times 0.027$
$∴ n \times 96500 \times 0.027 = 10800 C$
$∴ n = 4$