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

A 5-ampere current is passed through a solution of zinc sulphate for $40 $ minutes. The amount of zinc deposited at the cathode is:

  1. $0.4065 g$
  2. $65.04 g$
  3. $40.65 g$
  4. $4.065 g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\because$ $W=z.i.t$=$\cfrac{E}{F}\times{i}.{t}$$=\left(\cfrac{65.38\times5\times40\times60}{2\times96500}\right)g$.


                                $=4.065 g.$

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

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $\left[ Ni{ \left( { NO } _{ 3 } \right)  } _{ 2 } \right]$ land chromium nitrate $\left[Cr{ \left( { NO } _{ 3 } \right)  } _{ 3 } \right]$ respectively. If $0.3g$ of nickel was deposited in the first cell, the common of chromium deposited is :
$(at. Wt. Of Ni=59, at. Wt. Of Cr=52)$

  1. $0.1g$
  2. $0.17g$
  3. $0.3g$
  4. $0.6g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Faraday's laws of electrolysis, the mass deposited is proportional to the equivalent weight. Equivalent weight of Ni is 59/2 = 29.5, and for Cr is 52/3 = 17.33. Mass of Cr = (Mass of Ni * Eq. Wt. Cr) / Eq. Wt. Ni = (0.3 * 17.33) / 29.5 = 0.176g.

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

A certain quantity of electricity when passed through solution of ${ AgNO } _{ 3 }$, ${ ZnSO } _{ 4 }$, ${ CrI } _{ 3 }$. If X moles of Cr are deposited at its cathode, how many moles of Ag and Zn are deposited at their respective cathodes.

  1. X, X

  2. 3X, 2X

  3. 3X, 1.5X

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Ag^++e^- \longrightarrow Ag$

$Zn^{2+}+2e^- \longrightarrow Zn$
$Cr^{3+}+3e^-\longrightarrow Cr$
Given that $x$ moles of $Cr$ is deposited at it's cathode means in case of $Ag$ it will be $3X$  and for $Zn$ it will be $\cfrac {3X}{2}$ moles deposits at cathode.

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

How long (approximate) should water be electrolysed by passing through $100$ amperes current so that the oxygen realised can completely burn $27.66\ g$ of diborane?


(Atomic weight of $B=10.8\ u$ )

  1. $0.8$ hours
  2. $3.2$ hours
  3. $1.6$ hours
  4. $6.4$ hours
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The combustion of diborane (B2H6 + 3O2 -> B2O3 + 3H2O) requires 3 moles of O2 per mole of B2H6. 27.66g of B2H6 is 1 mole (MW=27.66). Thus, 3 moles of O2 are needed. Electrolysis of water (2H2O -> 2H2 + O2) requires 4 moles of electrons per mole of O2. Total charge Q = 3 * 4 * 96500 C. Time = Q / I = 1158000 / 100 = 11580 seconds, which is approximately 3.2 hours.

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

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $[Ni(NO _{3}) _{2}$] and chromium nitrate $[Cr(NO _{3}) _{3}$] respectively.If 0.3 g of nickel was deposited in the first cell, the amount of chromium deposited is:
(at.wt of Ni=59, at. wt. of Cr=52)

  1. 0.1 g

  2. 0.17 g

  3. 0.3 g

  4. 0.6 g
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

On electrolysing a solution of dilute ${ H } _{ 2 }{ SO } _{ 4 }$ between platinum electrodes, the gas evolved at the anode is 

  1. ${ SO } _{ 2 }$
  2. ${ SO } _{ 3 }$
  3. ${ O } _{ 2 }$
  4. ${ H } _{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

During the electrolysis of dilute sulfuric acid using platinum electrodes, water is oxidized at the anode rather than sulfate ions, releasing oxygen gas (O2).

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 electrolyte is selected for the electroplating of metal article with silver?

  1. Argentocyanide

  2. Silver nitrate

  3. Silver chloride

  4. Cyanochloride
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sodium argentocyanide (Na[Ag(CN)2]) is used as the electrolyte for silver plating because it provides a very low concentration of Ag+ ions, which ensures a smooth and uniform deposit.

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.