Question

In an electrolysis of a metallic chloride 3.283 g of the metal (molar mass 197gmol−1) was deposited on the cathode by the passage of 4825C of electric charge.The charge number of metal ion is

Answer 1

Answer:

Explanation:m^+n + ne...........m

nF. 197

4825. 3.283

4825*197 = n* 96500*3.283

n = 3

Answer 2

Given:

• Mass of the metal = 3.283g
• The atomic mass of the metal = 197 g/mol
• Amount of electric charge deposited = 4825 C

To Find:

• The charge number of the metal ion.

Solution:

• We already know the standard equation for the electrolysis of metals.                

    ⇒ $m^{n+}$ + ne → m

• From the given data we get an equation,

    ⇒ 4825*197 = 3.283*nF

• The above formula is obtained using the chemical equation mentioned and the corresponding values where F = 96500)

   ⇒ n = $\frac{4825*197}{3.283*96500}$ = 950525/316809.5 = 3.0003

   ⇒ n = 3.0003

 

∴ The charge number of the metal ion is 3.003