Question

A current of 2 amp when passed for 5 hours a metal salt deposits salt deposits 22.2 g metal atomic 177 the oxi dation sate of metal in metal salt

Answer 1

Answer:

+3

Explanation:

W=E*I*t/96500

t=5*60*60

W=22.2

I=2

E=96500*22.2/2*5*60*60

E=59.5

oxidation number= 177/59.5= +3

Answer 2

Answer:

The oxidation state of metal in metal salt is 3.

Explanation:

From the faradays law, we have,

$w=\frac{ItE}{96500}$              (1)

Where,

w=weight of a metal deposited at an electrode

I=current flowing through an electrode

t=time during which the current flows

E=equivalent weight of the metal

From the question we have,

I=2A

t=5 hours=5×3600sec

w=22.2g

The atomic mass of the metal=177g

By substituting the required values in the equation (1) we get;

$22.2=\frac{2\times 5\times 3600\times E}{96500}$

$E=59.5$

And the equivalent weight of the metal is given as,

$E=\frac{M_a}{n}$         (2)

$M_a$=atomic mass of the metal

n=oxidation state of the metal

By substituting the required values in equation (2) we get;

$59.5=\frac{177}{n}$

$n=3$

Hence, the oxidation state of metal in metal salt is 3.