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 energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

Which of the following reactions is used to make a fuel cell?

  1. $Cd(s) + 2Ni(OH) _{3} (s) \rightarrow CdO(s) + 2Ni(OH) _{2} (s) + H _{2}O(l)$
  2. $Pb(s) + PbO _{2}(s) + 2H _{2} SO _{4}(aq) \rightarrow 2PbSO _{4}(s) + 2H _{2}O(l)$
  3. $2H _{2}(g) + O _{2}(g) \rightarrow 2H _{2}O(l)$
  4. $2Fe(s) + O _{2}(g) + 4H^{+} (aq) \rightarrow 2Fe^{2+}(aq) + 2H _{2}O(l)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Reaction used in fuel cell is
$2H _{2}(g) + O _{2}(g) \rightarrow 2H _{2}O (l)$

At anode: $[H _{2} \rightarrow 2H^{+} + 2e]\times 2$
At cathode: $O _{2} + 2H _{2}O + 4e\rightarrow 4OH^{-}$

Hence, option C is correct.

Multiple choice chemistry energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

Fuel cell involves which of the following reaction(s) ?

  1. $O _{2}(g) + 2H _{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq.)$ (at cathode)
  2. $O _{2}(g) + 2H _{2}O(l) + 4e^{-}\rightarrow 4OH^{-}(aq.)$ (at anode)
  3. $2H _{2}(g) + 4OH^{-}(aq.) \rightarrow 4H _{2}O(l) + 4e^{-}$ (at anode)
  4. $2H _{2}(g) + 4OH^{-}(aq.) \rightarrow 4H _{2}O(l) + 4e^{-}$ (at cathode)
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Question approach,

at anode- loss of electron takes place,
at cathode- gain of electron takes place,
in option B, at anode gain of electron is talking place and in option D at cathode, loss of elctron is taking place;
so, both these statements are wrong
option A and C are representing the Hydrogen fuel cell reactions
hence, both A and C are correct options. 

Multiple choice chemistry energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

Which of the following statements regarding fuel cell is (are) true?

  1. They have efficiency of about 60-70%

  2. There efficiency is less then conventional methods

  3. They do not cause pollution

  4. They can supply indefinite amount of energy until outside supply of reactants is maintained
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Fuel cells have efficiency of 60-70%. Reactants, products and catalyst used are non- toxic, thus, they are pollution-free. And they can supply indefinite amount of energy until outside supply of reactants like H$ _{2}$ and O$ _{2}$  is maintained.

Multiple choice chemistry energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

In a $H _2 - O _2$ fuel cell, combustion of hydrogen occurs to:

  1. remove adsorbed oxygen from electrode surface

  2. create potential difference between the two electrodes

  3. produce high purity water

  4. generate heat
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Combustion of hydrogen occur in ${ H } _{ 2 }-{ O } _{ 2 }$ fuel cell because of difference in potential between two electrode. So on one electrode oxidation occur while on another reduction.

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

How many electron volts make one Joule?

  1. $ 3.25 \times 10^{19} \mathrm{eV} $
  2. $ 6.25 \times 10^{18} \mathrm{eV} $
  3. $ 9.25 \times 10^{17} \mathrm{eV} $
  4. $ 1.25 \times 10^{20} \mathrm{eV} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since 1 eV = 1.6 * 10^-19 Joules, the number of eV in 1 Joule is 1 / (1.6 * 10^-19), which equals 6.25 * 10^18.

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
D 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
B 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 ${MnO} _3^{-}$.