Calculate the volume of O2 liberated at STP by passing 5A current for 193 sec through acidified water:

Question

Calculate the volume of O2 liberated at STP by passing 5A current for 193 sec through acidified water: (A) 56mL (B) 112mL (C) 158mL (D) 965mL

Tardigrade Admin · Accepted Answer

As per the first law of electrolysis, The mass deposited or volume liberated at the electrode is directly proportional to the charge passing between the electrode.

m = ZQ

Where

Z = \frac{E}{F}

(electrovalent equivalent)

Q = i \times t

.

mass of

O_{2}

= \frac{E \times i \times t}{F}

(Where

F = 96500 C

)

= \frac{8 \times 5 \times 193}{96500} g

= 0.08 g

∵

At

STP

, volume of

32 g O_{2} = 22400 m L

∴

The volume of

0.08 g O_{2}

= \frac{22400 \times 0 . 08}{32}

= 56 m l

Q. Calculate the volume of $O_{2}$ liberated at $STP$ by passing $5 A$ current for $193 sec$ through acidified water:

$56 m L$

$112 m L$

$158 m L$

$965 m L$

Q. Calculate the volume of O2​ liberated at STP by passing 5A current for 193sec through acidified water:

56mL

112mL

158mL

965mL

Q. Calculate the volume of $O_{2}$ liberated at $STP$ by passing $5 A$ current for $193 sec$ through acidified water:

$56 m L$

$112 m L$

$158 m L$

$965 m L$