Question

Find out weight of a substance deposite on cathode when 1.5 ampere current is passed through an aqueous solution for 10 minutes?
(Given:‐F=96500 , molarmass=100 , Bivalent element)​

Answer 1

Answer:

Half cell reduction reaction:-

X

+2

+2e

−

→X

(s)

.

→ Mole Ratio=

molesofe

−

molesofXformed

=

2

1

.

→ Moles of X produced=

96500Cmol

−1

IXt(C)

×

2

1

AtomicMass

Weight

=

96500Cmol

−1

1.5×1800(C)

×

2

1

----------[Weight→ mass deposited at cathode.]

∴

1.5×1800(C)

0.889g×96500Cmol

−1

×2

=AtomicmassofX

∴ Atomic mass of X=63.5g

Explanation:

Hope it helps.....

Answer 2

Answer:

answer annawer answer