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 applications of electrolysis electroplating extraction of metals by electrolysis

Which process occurs in the electrolysis of an aqueous solution of nickel chloride at nickel anode?

  1. $Ni\rightarrow Ni^{2+}+2e^{-}$
  2. $Ni^{2+}+2e^{-}\rightarrow Ni$
  3. $2CI^{-}\rightarrow 2CI _{2}+2e^{-}$
  4. $2H^{+}+2e^{-}\rightarrow H _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At a nickel anode in an electrolytic cell, the metal itself undergoes oxidation (Ni -> Ni2+ + 2e-) because nickel is an active electrode.

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

On the basic of information available from the reaction
$ 4AI + 3O _{2} \rightarrow 2AI _{2}O _{3}; \triangle G = -965$ kJ/mol of $O _{2}$
The minimum EMF required to carry out electrolysis of $ AI _{2}O _{3}$ is

  1. 0.833 V

  2. 2.5 V

  3. 5.0 V

  4. 1.67 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Delta G = -nFE. For the reaction 4Al + 3O2 -> 2Al2O3, n = 12 electrons. Delta G = -965 kJ/mol O2. Since 3 moles of O2 are used, total Delta G = 3 * -965 = -2895 kJ. E = -Delta G / nF = 2895000 / (12 * 96500) = 2.5 V.

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

Which of the following reaction taken place at anode during electroplating of metal with silver using sodium argentocyanide as an electrolyte?

  1. $Ag\quad -{ e }^{ \ _ }\longrightarrow { Ag }^{ \ + }$
  2. ${ Ag }^{ + }+{ e }^{ \ _ }\longrightarrow Ag$
  3. $Ag\quad -2{ e }^{ \ _ }\longrightarrow { Ag }^{ \ +}\quad +{ e }^{ \ - }$
  4. $2{ Ag }^{ + }+{ e }^{ \ _ }\longrightarrow Ag$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In electroplating with silver, oxidation reaction at anode takes place:


$Ag - e^-\rightarrow Ag^+$.

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

Which of the following reaction takes place at the cathode during electroplating of metal with silver using sodium argentocyanide as an electrolyte?

  1. ${ Ag }^{ + }+{ e }^{ \ _ }\longrightarrow Ag$
  2. $Ag \longrightarrow { Ag }^{ \ _ } + { e }^{ \ _ }$
  3. $2{ Ag }^{ + }+{ e }^{ \ _ }\longrightarrow Ag$
  4. $Ag^+\longrightarrow { Ag }^{ \ _ } +{ e }^{ \ _ }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In electroplating with silver,
reaction at cathode:
$Ag^+ +e^-\rightarrow Ag$.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis
In an electroplating experiment with a $Cu^{2+}$ solution, $10.0$ amp is applied for $965\ sec$. How many moles of $Cu$ will be plated?
  1. $0.05\ \text{moles}$
  2. $0.1\ \text{moles}$
  3. $0.001\ \text{moles}$
  4. $0.005\ \text{moles}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$w = {Z\times I\times t}$

$w = \dfrac{E\times I\times t}{96500}$

$w = \dfrac{Molecular\ weight \times I\times t}{2\times 96500}$                                                 [$\therefore E= M/n$]

n=2

$n = \dfrac{w}{M} = \dfrac{I\times t}{2\times 96500}$

$ n=$ $\dfrac{10\times 965}{2\times 96500} = 0.05\ moles$

Hence, option A is correct.
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

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
A 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

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.