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 electrical cells patterns and properties of metals chemistry

The following electrochemical cell is taken $Cu| Cu^{2+}(aq)|| Ag^{+}(aq))| Ag$ and has emf $E _{r}>0$ by which of the following actions $E _{1}$ increases?

  1. Adding $NH _{3}$ to the cathodic chamber
  2. Adding $HCl$ to the cathodic chamber
  3. Adding $AgNO _{3}$ to the anodic chamber
  4. Adding $NH _{3}$ to the anodic chamber
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice electrical cells patterns and properties of metals chemistry

Find out ${E} _{cell}$ of following electrochemical cell (${E} _{{Br} _{2}/{Br}^{-}}=1.09V$)
$Pt(s)\mid {Br} _{2}(l)\mid{Br}^{-}(0.01M)\mid\mid{H}^{+}(0.01M)\mid{H} _{2}(g)(1bar)\mid Pt(s)$

  1. $-1.32V$
  2. $+1.09V$
  3. $-2.15V$
  4. $-1.92V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice electrical cells patterns and properties of metals chemistry

The following electrochemical cell has been set up: $Pt(s)|Fe^{3+}, Fe^{2+}(a=1)||Ce^{4+}, Ce^{3+}(a=1)|Pt(s); E^{\circ}(Fe^{3+}|Fe^{2+})= 0.77V$; $E^{\circ}(Ce^{4+}|Ce^{3+})= 1.61V$. If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?

  1. Ce electrode to Fe electrode, decrease

  2. Ce electrode to Fe electrode, increase

  3. Fe electrode to Ce electrode, decrease

  4. Fe electrode to Ce electrode, increase
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The cell potential is E_cell = E_cathode - E_anode = 1.61 - 0.77 = 0.84V. Since E_cell > 0, the reaction is spontaneous. Electrons flow from the anode (Fe) to the cathode (Ce). Current flows from the Ce electrode to the Fe electrode. As the reaction proceeds, the concentrations change, moving toward equilibrium, which causes the voltage and current to decrease.

Multiple choice electrical cells patterns and properties of metals chemistry

Consider a spontaneous electrochemical cell containing $Cd, Cd^{2+}, Ag^{+}$, and $Ag$.
The reduction potential of $Cd$ is $-0.403\ V$ and $Ag$ is $0.799\ V$.

What is the balanced equation for the reaction that is occurring, and what is the electrochemical cell voltage?

  1. $Ag^{+} + Cd \rightarrow Cd^{2+} + Ag; 1.19\ V$
  2. $2Ag^{+} + Cd \rightarrow Cd^{2+} + 2Ag; 1.20\ V$
  3. $Ag^{+} + Cd \rightarrow Cd^{2+} + Ag^{+}; 0.40\ V$
  4. $2Ag^{+} + Cd \rightarrow Cd^{2+} + 2Ag; 2,30\ V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have,
$E^0$ of Cd = -0.403V ( oxidation will favour)
$E^0$ of  Ag = 0.799V (reduction will favour)

So, the net reaction becomes:

$2 Ag^+ + Cd \rightarrow Cd^{2+} + 2Ag$

and $E^0 _{cell}  = 0.799 - (-0.403) = 1.202 V$
Multiple choice electrical cells patterns and properties of metals chemistry

An electrochemical cell consists of?

  1. A cathode, anode, electrolyte, wire and two compartments.

  2. Cathode, anode, and wire

  3. Two compartments that conduct electricity.

  4. A positive and negative side.

  5. A cathode and an anode.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

An electrochemical cell consists of a cathode, anode, electrolyte, wire and two compartments. An electrochemical cell is a device used to study chemical reactions electrically.

Multiple choice electrical cells patterns and properties of metals chemistry

Consider the reaction;
$Cl _2(g)+2Br^-(aq)\rightarrow 2Cl^-(aq)+Br _2$
The emf of the cell when
$[Cl^-]=[Br _2]=[Br]=0.01 M$ and $Cl _2$ gas at 1 atm pressure will be: ($E^0$ for the above reaction is = 0.29 volt)

  1. 0.54 volt

  2. 0.35 volt

  3. 0.24 volt

  4. -0.29 volt
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice electrical cells patterns and properties of metals chemistry

Foe the electrochemical cell:
$Zn\left( s \right) |{ Zn }^{ 2+ }\left( aq \right) \parallel { Cl }^{ - }\left( aq \right) |{ Cl } _{ 2 }\left( g \right) |Pt\left( s \right) $
Given : ${ E } _{ { Zn }^{ 2+ }/Zn }^{ o }=-0.76\ Volt$
             ${ E } _{ { Cl }^{ - }/{ Cl } _{ 2 }\left( g \right)  }^{ o }=-1.36\ Volt$
From these data one can deduce that:

  1. $Zn + Cl _{2} \rightleftharpoons Zn^{2+} + 2Cl^{-}$ is a non-spontaneous reaction at standard conditions.
  2. $Zn^{2+} + 2Cl^{-} \rightleftharpoons Cl _{2} + Zn$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 2.12\ volt.$
  3. $Zn + Cl _{2} \longrightarrow Zn^{2+} + 2 Cl^{-}$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 2.12\ volt.$
  4. $Zn + Cl _{2} \longrightarrow Zn^{2+} + 2 Cl^{-}$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 0.60\ volt.$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$Cl=\to zn^{2+} +2e^- \to Zn$
$2Cl\to Cl _2^- +2e^-$
for sponlareous reaction $E^o$ cell $>$ zero
$\Delta G <$ zero
$Zn^{3+}+2Cl^{-}\to Cl _2+Zn$ 
$Zn \to Zn^{2+} +2e^-$
$Cl _2 \to 2e^- \to 2Cl^- $(cathode)
$Zn +Cl _2 \to Zn^{3+}+2Cl^- ....(1)$
$E^o _{RP}$ (cathode) $=0.76$
$E^o $ cell $=E^o _{OP}+E^o _{RP}$
$=2.12$
as $E^o > 0$
$\Delta G < $ zero
so reaction $(1)$ is sponlareous at $2.12$
Multiple choice electrical cells patterns and properties of metals chemistry

The electrochemical cell shown below is a concentration cell.
$M|{ M }^{ 2+ }$ (saturated solution of a sparingly soluble salt, $M{X} _{2})\parallel {M}^{2+}(0.001 mol{dm}^{-3})| M$. 
The emf of the cell depends on the difference in concentrations of ${M}^{2+}$ ions at the two electrodes.
The emf of the cell at $298K$ is $0.099V$
The solubility product (${K} _{sp}:{mol}^{3}{dm}^{-9}$) of ${MX} _{2}$ at $298K$ based on the information available for the given concentration cell is: (take $2.303\times R\times 298/F=0.059V$)

  1. $1\times {10}^{-15}$
  2. $4\times {10}^{-15}$
  3. $1\times {10}^{-12}$
  4. $4\times {10}^{-12}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice bio-chemistry biological oxidation classical idea of redox reactions oxidation reduction redox reaction

How many moles of electrons are involved in the reduction of one mole of $MnO^- _4$ ion in alkaline medium to $MnO^- _3$?

  1. $2$
  2. $1$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Reaction takes place as follows:
$Mn{ O } _{ 4 }^{ - }+{ H } _{ 2 }O+{ 2e }^{ - }\rightarrow Mn{ O } _{ 3 }^{ - }+2O{ H }^{ - }$
So, 2 electrons are involved in reduction of 1 mole of ${MnO} _4^{-}$ ion in alkaline medium to ${MnO} _3^{-}$.
Multiple choice bio-chemistry biological oxidation classical idea of redox reactions oxidation reduction redox reaction

In acid medium, the standard reduction potential of $NO$ converted to ${ N } _{ 2 }O$ is $1.59 V$. Its standard potential in alkaline medium would be:

  1. $-1.59 V$
  2. $-0.764 V$
  3. $0.764 V$
  4. $0.062 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$NO \longrightarrow N _2O$      $E^o= 1.59V$

The reaction in acidic medium,
$2NO+2H^+ \longrightarrow N _2O+H _2O$;   $E^o=1.59$
In basic medium,
$2NO+H _2O \longrightarrow N _2O+2OH^-$;   Let $E^o=x$
Now, in alkaline medium,
$H _2O \rightleftharpoons H^++OH^-$
$\Rightarrow K=10^{-14}$
or, $2H _2O \rightleftharpoons 2H^++2OH^-$,   $K= 10^{-28}$
We know, $E^o= \cfrac {0.059}{n}\log K$
$=\cfrac {0.059}{2}\log 10^{-28}= 0.826$
For standard potential,
$E^o _{acidic}+E^o _{basic}=E^o _{cell}$
$\Rightarrow 1.59+x= 0.826$
$\Rightarrow x= 0.764$

Multiple choice botany cell theory, cell specialization, and cell replacement number and shape of cells unicellular and multicellular organisms diversity and classification in animals

The maximum work that a standard Daniell cell can do before it stops working is
$[1 \,F = 96487\, C\, mol^{-1}\, and\, E^o = 1.1 V]$

  1. $212.271\, kJ \,mol^{-1}$
  2. $21.227\, kJ \,mol^{-1}$
  3. $721.221 \,kJ \,mol^{-1}$
  4. $71.222 \,kJ \,mol^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The maximum work (Gibbs free energy change) is calculated as W = nFE. For a Daniell cell, n=2. W = 2 * 96487 * 1.1 = 212271.4 J/mol, which is 212.271 kJ/mol.

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

The ratio of volumes of $H _2$ and $O _2$ liberated on electrolysis of water is

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

$H _{2}O \rightarrow H^{+} +OH^{-}$
Cathode  :  4$OH^{-} \rightarrow 2H _{2}O +O _{2}+4e^{-} $
Anode :  4$H^{+} +4e^{-} \rightarrow 2H _{2} $
The ratio of volumes of H$ _{2}$ and O$ _{2}$ is $2 : 1$.

Multiple choice chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

The EMF of the cell: Ag, AgCl in $0.1 M - KCl \parallel$ satd. $ NH _{4}NO _{3}\parallel 0.1 M - Ag NO _{3},$ Ag is 0.42 V at $ 25^{\circ}C.$ 0.1 M-KCl is 50% dissociated and $0.1 M - Ag NO _{3}$ is 40% dissociated. The solubility product of $ AgCl $ is $ (2.303 RT/F = 0.0)$

  1. $1.0 \times 10^{-10} $
  2. $ 2.0 \times 10^{-9} $
  3. $ 1.0 \times 10^{-9}$
  4. $ 2.0 \times 10^{-10} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the Nernst equation and the given dissociation constants, one can calculate the concentration of Ag+ ions and Cl- ions to find the solubility product Ksp = [Ag+][Cl-].