Tag: electrolysis and its applications

Questions Related to electrolysis and its applications

During electrolysis of fused calcium hydride, the hydrogen is produced at:

  1. cathode

  2. anode

  3. hydrogen is not liberated at all

  4. $H _2$ produced reacts with oxygen to form water


Correct Option: B
Explanation:

$CaH _2\, \xrightarrow {Electrolysis}\, Ca^{2+}\, +\, 2H^{-}$


At cathode (reduction) :
$Ca^{2+}\, +\, 2e^-\, \rightarrow\, Ca$

At anode (oxidation) :
$2H^{-}\, \rightarrow\, H _2\, +\, 2e^-$

At the anode, hydrogen is produced. 

Hence, option $B$ is correct.

The gas evolved at the anode when $K _2SO _4\, (aq)$ is electrolyzed between Pt electrodes is:

  1. $O _2$

  2. $H _2$

  3. $SO _2$

  4. $SO _3$


Correct Option: A
Explanation:

Here,
Electrolysis of $H _2O$ takes place:

$2H _2O\, \rightarrow\, O _2\, +\, 4H^{\oplus}\, +\, 4e^-$

and $O _2$ evolved.

Which of the following statements is/are correct ?

  1. The electrolysis of concentrated $H _{2}SO _{4}$ at $0 - 5^{\circ}C$ using a Pt electrode produces $H _{2}S _{2}O _{8}$

  2. The electrolysis of a brie solution produces $NaClO _{3}$ and NaClO.

  3. The electrolysis of a $CuSO _{4}$ solution using Pt electrodes causes the liberation of $O _{2}$ at anode and the deposition of copper at cathode.

  4. All electrolytic reactions are redox reactions.


Correct Option: A,C,D
Explanation:

Electrolysis of concentrated $H _{2}SO _{4}$ at $0 - 5^{\circ}C$ using Pt electrode produces $H _{2}S _{2}O _{8}$.

At anode : $2H _{2}SO _{4}\, \rightarrow\, H _{2}S _{2}O _{8}\, +\, 2H^{\oplus}\, +\, 2e^{-}$

Statement (b) is wrong. Electrolysis of brine (aq. NaCl) gives $H _{2}(g)$ at cathode and $Cl _{2}(g)$ at anode.

Statements (c) and (d) are factual statements.

Hence, A, B and D are correct options.

When an aqueous solution of $CaCl _{2}$ is electrolyzed using inert electrodes, which of the following is(are) true ?

  1. Calcium deposits on cathode

  2. Calcium deposits an anode

  3. Chlorine is liberated on anode

  4. Calcium hydroxide precipitates near cathode on prolonged hydrolysis.


Correct Option: C,D
Explanation:

During electrolysis of aqueous $CaCl _2$

We have,
At anode: $2Cl^- \rightarrow Cl _2 + 2e^-$
At anode: $2H^+ + 2e^- \rightarrow H _2$
Electrolyte = $Ca^{2+} + OH^-$

Which of the following does not evolve oxygen at anode when the electrolysis is carried out of:

  1. Dilute $H _{2}SO _{4}$ with $Pt$ electrodes

  2. Fused sodium hydroxide with $Pt$ electrodes

  3. Acidic water with $Pt$ electrodes

  4. Dilute sulphuric acid using $Cu$ electrodes


Correct Option: D
Explanation:

$E^{\ominus} _{ox}$ of $Cu$ =$ - 0.34 $
Using copper electrodes, copper will oxidize : 
$Cu\, \rightarrow\, Cu^{2+}\, +\, 2e^{-}$
So oxygen is not evolved.

A dilute aqueous solution of sodium fluoride is electrolyzed, the products at the anode and cathode are:

  1. $O _{2},\, H _{2}$

  2. $F _{2},\, Na$

  3. $O _{2},\, Na$

  4. $F _{2},\, H _{2}$


Correct Option: A
Explanation:

Ion with higher potential will go to corresponding electrode and discharge.
So,
$NaF$:
Cathode : $2H^{\oplus}\, +\, 2e^{-}\, \rightarrow\, H _{2}$
Anode : $4\overset{\ominus}{O}H\, \rightarrow\, O _{2}\, +\, 2H _{2}O\, +\, 4e^{-}$

A solution of sodium sulphate was electrolyzed using some inert electrode. The products at the electrodes are:

  1. $O _2,\, H _2$

  2. $O _2,\, Na$

  3. $O _2,\, SO _2$

  4. $O _2,\, S _2O _8^{2-}$


Correct Option: A
Explanation:

Since the reduction potential of $H _2O$ is greater than reduction potential of $Na^{\oplus}\, so\, H _2O$ undergoes reduction to give
$H _2(g)$ at cathode.

$\displaystyle H _2O\, +\, e^-\, \rightarrow\, \overset {\ominus}O H +\, \frac{1}{2} H _2(g)$

Similarly, the oxidation potential of $H _2O$ is greater than the oxidation potential of $SO _4^{2-}$ ion. So, $H _2O$ undergoes oxidation to give $O _2(g)$ at anode.

$\displaystyle H _2O\, \rightarrow\, 2H^{\oplus}\, +\, 2e^-\, +\, \frac{1}{2} O _2(g)$

Which one is wrong if electrolysis of $CH _3COONa (aq)$ is made using $Pt$ electrodes?

  1. $pH$ of solution increases

  2. Molar ratio of gases at anode and cathode is $ 3 : 1.$

  3. $[CH _3COO^{\ominus}]$ in solution decreases.

  4. The molar ration of gases at anode and cathode is $2 : 1.$


Correct Option: D
Explanation:

Sodium acetate undergoes electrolysis to form ethane gas, carbondioxide and hydrogen gas.

Hydrogen is liberated at cathode and ethane at anode.

Select the wrong statement.

  1. The electrolysis of molten $CaH _2$ liberates $H _2$ at cathode

  2. During the discharge of lead storage battery, sulphuric acid is consumed.

  3. Sulphur acts as a polymerizing agent in the vulcanization of rubber.

  4. Galvanization of iron denotes coating with Zn.


Correct Option: A

In Hall's process, in the production of Al, carbon is used as the anode material. The reactions are:

  1. Anode: $Al^3+3e^-\, \rightarrow\, Al$

    Cathode: $C(s)+2O^{2-}\, \rightarrow\, CO _2(s)\, +\, 4e^-$

  2. Anode: $C(s)+2O^{2-}\, \rightarrow\, CO _2(s)\, +\, 4e^-$

    Cathode: $Al^3+3e^-\, \rightarrow\, Al$

  3. Anode: $Al\, \rightarrow\, Ae^{3+}\, +\, e^-$

    Cathode: $CO _2\, +\, 4e^-\, \rightarrow\, C(s)\, +\, 2O^2$

  4. None of the above


Correct Option: B
Explanation:

In the production of Al (in Hall's process), reactions are:
Cathode:
$Al^{3+}\, +\, 3e^-\, \rightarrow\, Al$
Anode:
$2O^{-2}\, \rightarrow\, O _2\, +\, 4e^-$
$C\, +\, O _2\, \rightarrow\, CO _2$