Chemistry · Physics

Electrochemistry and Current Electricity

224 Questions

Review fundamental concepts of electrochemistry and current electricity through these practice questions. The set includes numerical problems on electrolysis, drift velocity, and electrode potentials. These topics frequently appear in engineering, CSIR NET, and state PSC prelims examinations for thorough preparation.

Electrolysis calculationsElectrode potentialNernst equationDrift velocityFuel cellsMolar conductivity

Electrochemistry and Current Electricity Questions

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

In the electrolysis of $Na _{2}SO _{4}$ solution using inert electrode

a) the anodic reaction is

$2H _{2}O\rightarrow O _{2}(g)+4e^{-}+4H^{+}$

b)$H _{2}(g)$ and $O _{2}(g)$ is produced in a molar ratio of 2:1

c) 23 grams of sodium is produced at the cathode

d) the cathode reaction is $Na^{+}+e^{-}\rightarrow Na$

  1. a and b are correct

  2. c,d are correct

  3. only c is correct

  4. all are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discharge potential of $ { H }^{ + } $ ions is lower than $ { Na }^{ + } $ ions. Hence $ { H }^{ + } $ ions will be liberated at cathode in preference to $ { Na }^{ + } $ ions.
The discharge potential of $ { OH }^{ - } $ ions is lower than $ { SO } _{ 4 }^{ 2- } $ ions. Hence $ { OH }^{ - } $ ions will be liberated at anode (in the form of $ { O } _{ 2 } $ molecule) in preference to $ { SO } _{ 4 }^{ 2- } $ ions. This is the electrolysis of water. Hence $ { H } _{ 2 } $ and $ { O } _{ 2 } $ are produced in the molar ratio of 2:1 which is same as that present in water molecule.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

During electrolysis of aqueous $CuBr _{2}$ using Pt electrode:

  1. $Br _{2}(g)$ is evolved at anode.

  2. Cu(s) is deposited at cathode
  3. $Br _{2}(g)$ is evolved at anode and $H _{2}(g)$ at cathode
  4. $H _{2}(g)$ is evolved at anode
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Aqueous $CuBr _{2}$ :
At cathode:

$Cu^{2+}\, +\, 2e^{-}\, \rightarrow\, Cu$;  
At anode :
$2Br^{\ominus}\, \rightarrow\, Br _{2}\, +\, 2e^{-}$

$Br _{2}(g)$ is evolved at anode and Cu(s) is deposited at cathode.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

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.
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

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.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

During electrolysis of aqueous $CaCl _2$

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

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

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. $F _{2},\, H _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

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. $O _2,\, S _2O _8^{2-}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

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 $2 : 1.$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Hydrogen is liberated at cathode and ethane at anode.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

During the electrolysis of an aqueous solution of $HCOOK$, the number of gases obtained at cathode, anode, and a total number of gases are:

  1. $1, 2, 3$
  2. $1, 2, 2$
  3. $2, 1, 3$
  4. $2, 1, 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At cathode:
$\displaystyle H _2O\, +\, e^-\, \rightarrow\, \overset{\ominus}O H\, +\, \frac{1}{2} H _2\, (one)$
At anode:
$2HCOO^{\ominus}\, \rightarrow\, 2CO _2\, +\, H _2\, (two)$
$H _2\, and\, CO _2$ at anode and $H _2$ at cathode

With the total number of gases $2$

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

The products formed when an aqueous solution of $NaBr$ is electrolysed in a cell having inert electrodes are :

  1. $Na$ and $Br _{2}$
  2. $Na$ and $O _{2}$
  3. $H _{2}, Br _{2}$ and $NaOH$
  4. $H _{2}$ and $O _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$NaBr\rightleftharpoons Na^{+} + Br^{-}$
$2H _{2}O + 2e\rightarrow H _{2} + 2OH^{-}$
$Na^{+} + OH^{-} \rightarrow NaOH$ At cathode
$Br^{-}\rightarrow Br + e^{-}$
$Br + Br \rightarrow Br _{2}$ At anode
So the products are $H _{2}$ and $NaOH$ (at cathode) and $Br _{2}$ (at anode).

Hence, option C is correct option.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

A solution containing one mole per lite of each $Cu(NO {3}) _{2}; AgNO _{3}; Hg _{2}(NO _{3}) _{2}$ and $Mg(NO _{3}) _{2}$ is being electrolysed by using inert electrodes. The values of standard electrode potentials (reduction potentials) are $Ag/ Ag^{+} = 0.80\ volt, 2Hg/ H _{2}^{2+} = 0.79\ volt, Cu/ Cu^{2+} = + 0.24\ volt, Mg/ Mg^{2+} = -2.37\ volt$. With increasing voltage, the sequence of deposition of metals on the cathode will be___________.

  1. $Ag, Hg, Cu$
  2. $Cu, Hg, Ag$
  3. $Ag, Hg, Cu, Mg$
  4. $Mg, Cu, Hg, Ag$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Higher the standard reduction potential, higher will be the tendency to undergo reduction so, more fast will be its deposition on electrode (cathode) .

$E^o _{Ag/Ag^+}=0.80V\ E^o _{2Hg/Hg _2^{2+}}=0.79V\quad \downarrow (Reduction\quad potential\quad decreases)\ E^o _{Cu/Cu^{2+}}=0.24V$    
Order of deposition= $Ag> Hg> Cu$

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

The emf of a cell containing sodium/ copper electrodes is $3.05\ V$, if the electrode potential of copper electrode is $+0.34\ V$, the electrode potential of sodium is:

  1. $-2.71\ V$
  2. $+ 2.71\ V$
  3. $-3.71\ V$
  4. $+3.71\ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$E _{cell}^{\circ}=E _{cathode}^{\circ}-E _{anode}^{\circ}$
$\Rightarrow3.05=0.34-(-2.71)$
$=3.05$
$\therefore$ Electrode potential of Sodium is $-2.71V$