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 applications of electrolysis electroplating extraction of metals by electrolysis

In the electrolysis of $CuCl _{2}$ solution, the mass of cathode increased by $6.4\ g$. What occurred at copper anode?

  1. $0.224$ litre of $Cl _{2}$ was liberated
  2. $1.12$ litre of oxygen was liberated
  3. $0.05\ mole\ Cu^{2+}$ passed into the solution
  4. $0.1\ mole\ Cu^{2+}$ passed into the solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the electrolysis process,

reduction takes place at cathode and oxidation takes place at anode
so, if we increase the mass on cathode then same number of moles get oxidised at anode and go into the solution,
so, moles of Cu passed into the solution = $\dfrac{6.4}{63.5} = 0.1 mole$

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

Which is correct about silver plating?

  1. Anode - pure $Ag$

  2. Cathode - object to be electroplated
  3. Electrolyte - $Na[Ag(CN) _{2}]$

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In silver plating, the object to be plated (e.g., a spoon) is made from the cathode of an electrolytic cell. 

The anode is a bar of silver metal, and the electrolyte (the liquid in between the electrodes) is a solution of silver cyanide, $AgCN$, in water.

When a direct current is passed through the cell, positive silver ions ($Ag^+$) from the silver cyanide migrate to the negative anode (the spoon), where they are neutralized by electrons and stick to the spoon as silver metal.

Hence, option D is correct.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

When water is electrolysed, hydrogen and oxygen gases are produced. If $1.008\ g$ of $H _{2}$ is liberated at cathode, what mass of $O _{2}$ is formed at the anode?

  1. $32\ g$
  2. $16\ g$
  3. $8\ g$
  4. $4\ g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac {W _{1}}{W _{2}} = \dfrac {E _{1}}{E _{2}}$
$\dfrac {1.008}{W _{2}} = \dfrac {1.008}{8}$
$\therefore W _{2} = 8\ g$
where, $E _{1}$ and $E _{2}$ are equivalent masses of hydrogen and oxygen respectively.

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

$Zn\left( s \right) \left|\ Zn{ { \left( CN \right)  } } _{ 4 }^{ 2- }\ \left( 0.5\ M \right) ,{ CN }^{ - }\left( 0.01 \right)  \right| \left|\ Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 }^{ 2+ }\ \left( 0.5\ M \right) ,{ NH } _{ 3 }\left( 1\ M \right)  \right|\ Cu\left( s \right) $
Given: ${ K } _{ f }$ of $Zn{ { \left( CN \right)  } } _{ 4 }^{ -2\  }=\ { 10 }^{ 16 }$, $\quad \quad \quad$ ${ K } _{ f }$ of $Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 }^{ 2+ }\ =\ { 10 }^{ 12 }$
$\displaystyle \quad \quad \ \ { E } _{ Zn|{ Zn }^{ -2 } }\ =\ 0.76V\ ;\ { E } _{ { Cu }^{ +2 }|Cu }\ =\ 0.34V\ ,\ \dfrac { 2.303RT }{ F } =0.06$
The emf of above cell is:

  1. $1.22\ V$
  2. $1.10\ V$
  3. $0.98\ V$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Calculate the mass of Ag deposited at cathode when a current of 2A was passed through a solution of $Ag{ NO } _{ 3 }$ for 15 min.
(Given : Molar mass of $Ag = 108\ g\ { mol }^{ -\ 1 }$ $\ 1F=96500\ C\ { mol }^{ -1 }$).

  1. $3.015\ g$
  2. $2.015\ g$
  3. $4.2\ g$
  4. $3.1\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

Molar Mass of Ag = 108 g/mol

$1F = 96500\ C mol^{−1}$

Reaction at cathode = $Ag + e^-  \rightarrow   Ag(s)$ 

$w = Zlt$

Where, w = Mass deposited at cathode

Z = electrochemical constant

I = current

t = time

Now I = 2amp

$t = 15\ min = 15\times 60 = 900\ seconds$

Z = Eq. wt of substance $/ 96500 = 108/96500$ 

So,

$w = \dfrac{108}{96500} \times 900 \times 2 $

= $2.015g$

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

The electrochemical equivalent of silver is $0.0011180g$. When an electric current of $0.5$ ampere is passed through an aqueous silver nitrate solution for $200sec$, the amount of silver deposited is:

  1. $1.1180g$
  2. $0.11180g$
  3. $5.590g$
  4. $0.5590g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to the data given when $1$ e is passed the amount of silver,

deposited is $0.0011180\,g$
$\therefore (0.5\times 200)c$ is passed
then $(0.0011180\times 100)g$
of silver gets deposited 
$=0.11180\,g$

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

$H _2(g)$ and $O _2(g)$ , can be produced by the electrolysis of water. What total volume (in $L$) of $O _2$ and $H _2$ are produced at $STP$ when a current of $30$ A is passed through a $K _2SO _4\, (aq)$  solution for 193 minutes?

  1. 20.16

  2. 40.32

  3. 60.48

  4. 80.64
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total charge passed is Q = I * t = 30 A * (193 * 60 s) = 347400 C. The moles of electrons transferred are n(e-) = Q / F = 347400 / 96500 = 3.6 mol. From the electrolysis of water, 2 moles of electrons produce 1 mole of gas total (0.5 mol O2 and 1 mol H2 per 2 moles of electrons, meaning 3 moles of total gas per 4 moles of electrons; specifically, 4 e- give 1 mole O2 (22.4 L) and 2 moles H2 (44.8 L)). Scaling by 3.6 mol of electrons gives (3.6 / 4) * 3 * 22.4 = 60.48 L total gas at STP.

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

In $Ag$ - $CuSO _{4}$ cell, silver electrode will serve as:

  1. anode

  2. cathode

  3. both anode and cathode

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In $Ag - CuSO _4$ cell, silver electrode will serve as cathode, as reduction occurs here and silver metal is deposited. Copper being high in the reactivity series than silver, undergoes oxidation, and silver ions undergo reduction.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Electrolysis of a solution of $Mn{ SO } _{ 4 }$ in aqueous sulphuric acid is a method for the preparation of $MnO _ 2$. Passing a current of $27A$ for $24$ hours gives $1kg$ of $MnO _2$. The current efficiency in this process is:

  1. $100$%
  2. $95.185$%
  3. $80$%
  4. $82.951$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reaction taking place on the given reaction are,

$Mn^{2+}+2H _2O\rightarrow MnO _2+2H^{+}+H _2$

Amount of current is given by,
$m= Z \times I \times t$

$I=m\dfrac{F\times x}{t\times M}$

    $=1000g\times{\dfrac{96500\times 2}{24\times 60 \times 60 \times 86.9}}$

    $\implies I=25.70A$

Now current efficiency$=\dfrac{25.7}{27} \times 100=95.185$ %

Hence,option B is correct answer.

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

Consider the reaction : $Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$


What is the quantity of electricity in coulombs needed to reduced $1\ mol$ of $Cr _2O _7^{2-}$?

  1. $6 \times 10^6C$
  2. $5.79\times 10^5C$
  3. $5.25 \times 10^5C$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$

The valency factor for Chromium is $6$.

Applying faraday's first law-
$ \dfrac{Q} {F} =mole \times V.f.$

$  Q            =mole \times V.f. \times F$ 

$  Q            =1 \times 6 \times 96500$

$  Q            =5.79\times 10^5\ C$
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Calculate the amounts of Na and chlorine gas produced during the electrolysis of fused $NaCl$ by the passage of 1 ampere current for 25 minutes. 

  1. 0.3565 gm and 0.55 gm

  2. 0.3565 gm and 0.66 gm

  3. 0.55 gm and 0.3565

  4. None of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$i= 1 \ amp$
$t= 25 \ min = 25 \times 60 \ sec$
Molar mass of $Na= 23 \ g \ mol^{-1}$
Molar mass of $Cl _2= 71 \ g \ mol^{-1}$
$Cl^-= \cfrac 12 {(12)}=35.5 \ g$
Formula used, $m= \cfrac {Molar \ mass}{n \times F} \times i \times t$
For $Na$, $n=1$
$m= \cfrac {23}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.3575 \ g$

For $Cl$, $n=1$
$m= \cfrac {35.5}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.5518 \ g$

$\therefore$ The amounts of $Na$ and $Cl$ gas produced are $0.3575 \ g$ and $0.5518 \ g$ respectively.
Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power.

Ion $Cl{ O } _{4}^{-}$ $I{ O } _{4}^{-}$ $Br{ O } _{4}^{-}$
Reduction potential $E^{\circ}$/V $E^{\circ}$= 1.19V $E^{\circ}$= 1.65V $E^{\circ}$= 1.74V
  1. $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$
  2. $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$
  3. $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$
  4. $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The reduction potential of the substance is the ability of the substance to be reduced.So as the reduction potential increases reducing ability of the  substance increases.It means oxidizing power of the substance (The ability of the substance to make the other substance to lose the electrons increases which noting but oxidizing power) increases.

So the decreasing order of oxidizing power is $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$.

Hence option $C$ is correct.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

An acidic solution of Cu+ salt containing 0.4 g of Cu+ is electrolysed until all the Cu is deposited. The electrolysis is continued for 7 more minutes with volume of the solution kept at 100 ml and the current at 1.2 amp. Calculate volume of the gases evolved at NTP during entire electrolysis( atomic weight of Cu= 63.6)

  1. H2 = 68.32 ; O2 = 85.23 ml

  2. H2 = 40.34 ; O2 =70.12 ml

  3. H2= 51.46 ; O2 = 90.12 ml

  4. H2=58.46 ; O2 = 99.68 ml
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a complex stoichiometry and electrolysis problem. After Cu+ is deposited, the electrolysis of water occurs (2H2O -> 2H2 + O2). Calculating moles of electrons from current and time, then applying Faraday's laws, leads to the volumes of H2 and O2 evolved.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Given below are the half cell reactions

${ Mn }^{ 2+ }+{ 2e }^{ - }\longrightarrow { Mn }  E=-1.81V$

$2\left( { Mn }^{ 3+ }+{ e }^{ - }\longrightarrow { Mn }^{ 2+ } \right) E=+1.51V$ will be

        




  1. 0.33V ; the reaction will not occur .

  2. 0.33V ; the reaction will occur.

  3. 2.69V ; the reaction will not occur.

  4. 2.69V ; the reaction will occur
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the cell potential, we combine the half-reactions. The overall reaction is 3Mn^2+ -> Mn + 2Mn^3+. The E_cell = E_cathode - E_anode. Using the given values, E_cell = 1.51 - 1.81 = -0.30V. Since E is negative, the reaction is non-spontaneous.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere?

  1. 26.805 h

  2. 153.61 h

  3. 107.22 h

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Solution:- (D) none of these
$2 {H} _{2}O \longrightarrow 2 {H} _{2} + {O} _{2}$
From the above reaction-
$1$ mole of ${H} _{2}O$ exchanges $2$ moles of electrons, then $2$ moles of ${H} _{2}O$ will exchange $4$ moles of electrons.
From Faraday's law of electrolysis,
$q = nF$
$\Rightarrow i \times t = nF \; \left( \because q = i \times t \right)$
$\Rightarrow t = \cfrac{nF}{i} = \cfrac{4 \times 96500}{2} = 193000 \; s = 53.61 \; hr$
Hence the time required is $53.61$ hours.