Chemistry · Physics

Electrochemistry and Current Electricity

272 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

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 $-2.71V$
Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

An acidic solution of $C{u^{2 + }}$ salt containing $0.4\,g$ of $C{u^{2 + }}$ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the current at $1.2$ amp. Find the volume of gases evolved at anode and Cathode at NTP during the entire electrolysis. 

  1. $78.20\,ml,\,\,99.78\,ml$
  2. $58.48\,ml,\,\,78.20\,ml$
  3. $98.56\,ml,\,\,58.24\,ml$
  4. $78.20\,ml,\,\,58.48\,ml$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Solution:- (C) $98.56 mL, \; 58.24 \; mL$
Assuming ${Cu}^{+2}$ salt to be $CuS{O} _{4}$, the reactions occuring at the electrodes would be-
At anode:-
${H} _{2}O \longrightarrow 2 {H}^{+} + \cfrac{1}{2} {O} _{2} + 2 {e}^{-}$
At cathode:- 
${Cu}^{+2} + 2 {e}^{-} \longrightarrow Cu$
Equivalent weight of ${Cu}^{+2} = 31.5 \; g$
$0.4 \; g$ of ${Cu}^{+2} = \cfrac{0.4}{31.5} = 0.0127 \; g$ equivalent
At the same time the oxygen deposited at anode.
Equivalent weight of oxygen $= 8 \; gm$
Mass of oxygen deposited $= 0.0127 \times \cfrac{8}{32} = 0.0031 \text{ mol}$
After the complete deposition of copper, the reactions would be-
At anode:-
${H} _{2}O \longrightarrow 2 {H}^{+} + \cfrac{1}{2} {O} _{2} + 2 {e}^{-}$
At cathode:-
$2 {H} _{2}O + 2{e}^{-} \longrightarrow {H} _{2} + 2 {OH}^{-}$

Given:-
$I = 1.2 \; A$
$t = 7 \text{ min} = 7 \times 60 = 420 \; s$
Amount of charge passed $= I \times t \ = 1.2 \times 7 \times 60 = 504 \; C$
Therefore,
Amount of oxygen liberated $= \cfrac{1}{96500} \times 504 = 0.00523 \; g$ equivalent $= \cfrac{8}{32} \times 0.00523 = 0.0013 \text{ mol}$
Amount of hydrogen liberated $= 0.00523 \; g$ equivalent $= \cfrac{1}{2} \times 0.00523 = 0.0026 \text{ mol}$
Now,
Gas evolved at anode $= {O} _{2}$
Total no. of moles of ${O} _{2}$ evolved $= 0.0031 + 0.0013 = 0.0044 \text{ mol}$
$\therefore$ Volume of gas evolved at anode $= 0.00447 \times 22400 = 98.56 \; mL$
Gas evolved at cathode $= {H} _{2}$
Total no. of moles of ${H} _{2}$ evolved $= 0.0026 \text{ mol}$
$\therefore$ Volume of gas evolved at cathode $= 0.0026 \times 22400 = 58.24 \; mL$
Hence the volume of gases evolved at anode and cathode at NTP during the entire electrolysis $98.56 \; mL$ and $58.24 \; mL$ respectively.

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

When an aqueous concentrate solution of lithium chloride is electrolysed using inert electrodes:

  1. $Cl _2$ is liberated at the anode
  2. $Li$ is deposited at the cathode
  3. as the current flows, $pH$ of the solution around the cathode remains constant
  4. as the current flows, $pH$ of the solution around the cathode increases
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

As chloride ion have more oxidation potential than hydroxide ion, so $Cl _2$ is liberated at the anode.
And reduction potential of hydrogen ion is more so hydrogen gas produces at cathode so as the current flows, pH of the solution around the cathode increases.

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

A solution containing $Na^{\oplus},\, NO _{3}^{\ominus},\, Cl^{\ominus}$, and $SO _{4}^{2-}$ ions, all at unit concentrations, is electrolyzed between nickel anode and plantinum cathode. As the current is passed through the cell :

  1. pH of the cathode increases

  2. Oxygen is the major product at anode

  3. Nickel is deposited at cathode

  4. Chlorine is the major product at anode
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

At cathode : Reduction of $Na^{\oplus}$ does not occur but reduction of $H _{2}O$ occurs to give $\overset{\ominus}{O}H$ and $H _{2} (g)$, so pOH decreases and pH increases. Hence, option A is correct


At anode : Oxidation of $Cl^{\ominus}$ ions occur to give $Cl _{2} (g)$. Likewise, oxidation of $NO^{\ominus} _{3}$ and $SO _{4}^{2-}$ does not occur, but oxidation of $H _{2}O$ occurs to give $H^{\oplus}$ ions and $H _{2}$ (g). So pH decreases at anode. Also, chlorine forms hence, option D is correct.

Multiple choice chemistry chemical effects of electric current applications of electrolysis electroplating extraction of metals by electrolysis

If $Zn/Zn^{+2}$ electrode is diluted 100 times then the change in electromotive force will be :-

  1. increases of 59 mV

  2. decreases of 59 mV

  3. increases of 29.5 mV

  4. decreases of 29.5 mV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the Nernst equation: E = E_cell - (0.059/n) * log(1/[Zn+2]). Diluting 100 times increases the concentration of the denominator in the log term, making the log term more negative, which increases the overall potential by (0.059/2) * log(100) = 0.059 V = 59 mV.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.

  1. $\text{1M CuSO} _4$ solution, $\text{1M CuCl} _2$ solution
  2. $\text{1M KCl}$ solution, $\text{1M Kl}$ solution
  3. $\text{1M AgNO} _3$ solution, $\text{1M Cu(NO} _3) _2$ solution
  4. $\text{1M KCl}$ solution, $\text{1M NaCl}$ solution
  5. $\text{1M CuBr} _2$ solution, $\text{1M CuSO} _4$ solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electrolysis of 1M KCl and 1M NaCl using inert electrodes both yield potassium or sodium ions remaining in solution while hydrogen gas is evolved at the cathode and chlorine gas at the anode, making their electrolysis products identical and indistinguishable.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Electrolysis rules of Faraday's states that mass depends on electrodes is proportional to:-

  1. $m \propto I^2$
  2. $m \propto Q$
  3. $m \propto Q^2$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Faraday's first law of electrolysis states that the mass of a substance deposited is directly proportional to the quantity of electricity (charge Q) passed through the electrolyte.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Cost of electricity for the production of $X$ litres of $H _2$ at $NTP$ at the cathode is Rs $X$, then cost of electricity for the production $X$ litres of $O _2$ gas at $NTP$ at the anode will be:


[Assume $1$ mole of electrons as one unit of electricity]

  1. $2X$
  2. $4X$
  3. $16X$
  4. $32X$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electrolysis of water gives:


${ H } _{ 2 }O\rightarrow { H } _{ 2 }+\tfrac { 1 }{ 2 } { O } _{ 2 }$


On electrolysis of water, hydrogen, and oxygen formed in the ratio $2:1$.
Since $X$ litres of ${ H } _{ 2 }$ is formed. Amount of ${ O } _{ 2 }$ formed will be $\tfrac { X }{ 2 } $. Since cost of production of electricity from $\tfrac { X }{ 2 } $ litres of ${ O } _2$ = Rs $X$

So, cost of production of electricity from $X$ litres of ${ O } _2$ = Rs $2X$.

So, the correct answer is option $A$.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

A solution of $CuSO _4$ is electrolysed for $7$ minutes with a current of $0.6A$. The amount of electricity passed is equal to:

  1. $4.2C$
  2. $2.6\times 10^{-3}F$
  3. $126C$
  4. $36C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Quantity of electricity passed $\displaystyle Q(C) = I(A) \times t(s)$
$\displaystyle Q(C)=0.6 \ A \times 7 \ min \times 60 \ s/min$
$\displaystyle Q(C)=252 \ C$
Number of faraday passed $\displaystyle = \dfrac {252 \ C}{96500 \ C/F}=2.6 \times 10^{-3} \ F$

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

During electrolysis of an aqueous solution of a salt, pH in the space near one of the electrodes is increased, which of the following salt solution was electrolysed?

  1. $KCl$
  2. ${ CuCl} _{ 2 }$
  3. ${ Cu(NO } _{ 3 }{ ) } _{ 2 }$
  4. ${ CuSO } _{ 4 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
As the cation $\left( {K}^{+} \right)$ in $KCl$ has lower electrode potential than $H$, hydrogen is liberated at cathode.

There is an accumulation of ${H}^{+}$ at one electrode, resulting in an increase in pH.

Hence, option A is correct.