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 oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

The life span of a Daniel cell may increased by:

  1. large Cu electrode
  2. lowering of CuSO$ _{4}$ concentration
  3. lowering of ZnSO$ _{4}$ concentration

  4. large zinc electrode
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a daniel cell at anode Zn gets converted to $Zn^{2+}$
$\therefore$ If the size of electrode is increases more zinc can be oxidized,
hence the life span of the cell may be increased

Hence, option d is correct.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

Estimate the cell potential of a Daniel cell having $1.0 M - Zn^{2+}$ and originally having $ 1.0 M - Cu^{2+}$ after sufficient ammonia has been added to the cathode compartment to make the $ NH _{3}$ concentration 2.0 M. Given: $ _{zn^{2+}|Zn}^{0} = 0.76 V, E _{Cu^{2+}|Cu}^{0} = + 0.34 V, K _{f}$ for $ Cu (NH _{3}) _{4}^{2+} = 1 \times 10^{12}.$

  1. 1.10 V

  2. 0.704 V

  3. 0.396 V

  4. 1.496 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

In a Daniel cell when $Cu$ and $Zn$ electrodes are connected current flows from:

  1. $Cu$ to $Zn$ within the cell
  2. $Cu$ to $Zn$ outside the cell
  3. $Zn$ to $Cu$ outside the cell

  4. all of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know outside the cell, electron flow from anode to cathode in the external current. Chemical energy is converted into electrical energy. The net reaction is the sum of two half-cell reactions.

For daniel cell, $Zn$ is at the anode and $Cu$ at the cathode.
The electron current flows from Zn to Cu.

So, the correct option is ( C )
Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

In the electrochemical cell $H _{2}(g),1atm|H^{+}(1M)||Cu^{2+}(1M)|Cu(s)$, which one of the following statements is true? 

  1. H$ _{2}$ is cathode, Cu is anode

  2. Oxidation occurs at Cu electrode
  3. Reduction occurs at H$ _{2}$ electrode
  4. H$ _{2}$ is anode, Cu is Cathode
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At anode oxidation takes place and at cathode reduction take place
$\therefore H _2$ is anode and Cu is cathode

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

The standard EMF of a Daniel cell at 298 K is $E _{1}$. When the concentration of $ZnSO _{4}$ is 1.0 M and that of $CuSO _{4}$ is 0.01 M, the EMF becomes $E _{2}$ at 298 K. The correct relationship between $E _{1}$ and $E _{2}$ is

  1. $E _{1} = E _{2}$
  2. $E _{2}=0$
  3. $E _{1} > E _{2}$
  4. $E _{1} < E _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the Nernst equation, E = E0 - (0.059/2) * log([Zn2+]/[Cu2+]). As [Cu2+] decreases from 1.0 M to 0.01 M, the log term increases, making the subtraction larger and thus E2 < E1.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

If a current of 1.0 A is drawn from the Daniel cell for 96.5 min, the cathode will gain in weight by (Cu = 63.5, Zn = 65.4)

  1. 1.905 g

  2. 1.962 g

  3. 3.81 g

  4. 3.924 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Faraday's law of electrolysis: mass = (I * t * M) / (n * F). I = 1.0 A, t = 96.5 * 60 seconds = 5790 s, M = 63.5, n = 2, F = 96500 C/mol. Mass = (1 * 5790 * 63.5) / (2 * 96500) = 1.905 g.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

Electrochemical Battery

An electrochemical battery is a device powered by oxidation and reduction reactions that are physically separated. So that the electrons must travel through a wire from the reducing agent to the oxidizing agent. 

The reducing agent loses electrons and is oxidized in a reaction that takes place at an electrode is called the anode. The oxidizing agent gains electrons and is reduced in a reaction that takes at an electrode is called the cathode.  

To maintain a net zero charge in each compartment, there is a limited flow of ions through a salt bridge. 
For example, in a car battery the reducing agent is oxidized by the following reaction, which involves a lead ( $Pb$ ) anode and sulfuric acid ( $H _2SO _4$). Lead sulfate ( $PbSO _4$ ), protons ( $H$ ), and electrons ( $e^-$) are produced.

At the cathode, the below reaction occurs :

$PbO _2+H _2SO _4+2H^++2e^-\rightarrow$ PbSO_4+2H_2O$
 
At the anode, the below reaction occurs :

$Pb+H_2SO_4 \rightarrow PbSO_4+2H^++2e^-$

Electrons are produced by a chemical reaction that takes place at the:

  1. anode.

  2. cathode.

  3. lead oxide electrode.

  4. oxidizer.

  5. salt bridge.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At the anode, oxidation of metal takes place;

oxidation is the process of removal of the electron;

So, the electrons that are produced in the chemical reaction take place at the anode.

Hence, option (A) is correct

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

For the cell $Zn(s)|Zn^{2+}(aq)||Cu^{+2}(aq)|Cu(s)$. The cell potential $E _{cell}$ can be increased.

  1. By increasing $[Cu^{+2}]$
  2. By increasing $[Zn^{+2}]$
  3. By decreasing $[Cu^{+2}]$
  4. By decreasing $[Zn^{+2}]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the Nernst equation, E = E0 - (0.059/2) * log([Zn2+]/[Cu2+]). To increase E, we must decrease the value of the log term, which means decreasing [Zn2+] or increasing [Cu2+].

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

In a Daniell cell:

  1. The chemical energy liberated during the redox reaction is converted to electrical energy

  2. The electrical energy of the cell is converted to chemical energy

  3. The energy of the cell is utilised in conduction of the redox reaction

  4. The potential energy of the cell is converted into electrical energy
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A galvanic cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals connected by a salt bridge, or individual half-cells separated by a porous membrane. Daniel cell is a galvanic cell.

Daniel cell has Zn as anode and Cu as cathode. The reactions of Daniel cell are,

At anode(Oxidation),

$Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}$

At cathode(Reduction),

$Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)$

Hence overall reaction is,

$Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)$

Here the chemical energy is liberated when the redox reaction takes place in the cell and it is converted into electrical energy.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

Which one is not correct for e.m.f. of a galvanic cell?

  1. $E _{cell} = E _{OP _{anode}} + E _{RP _{cathode}}$
  2. $E _{cell} = E _{OP _{LHS}} + E _{RP _{RHS}}$
  3. $E _{cell} =$ higher oxidation potential - lower oxidation potential
  4. $E _{cell} =$ lower oxidation potential - higher oxidation potential
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

The correct expressions for the standard cell potentials of a galvanic cell are as given below.
$E _{cell} = E _{OP _{anode}} + E _{OP _{cathode}}$
$E _{cell} = E _{OP _{LHS}} + E _{OP _{RHS}}$
$E _{cell} =$lower oxidation potential $-$ higher oxidation potential.
Thus, only option D is correct and options A to C are incorrect.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

Identify the true statement regarding Daniel cell:

  1. Zinc ions flows across salt bridge

  2. ${K}^{+}$ ions move from salt bridge to $Cu/{Cu}^{+2}$ half cell

  3. Oxidation takes place at copper electrode

  4. Flow of current takes place from copper electrode to zinc electrode
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Daniell cell is  perform the spontaneous redox reaction between zinc and cupric ions to produce an electric current. It consists of two half-cells. The  left half cell contains a zinc metal electrode dipped in $ZnSO _4$ solution.The half right half cell consists of copper metal electrode in a solution $CuSO _4$. The half-cells are joined by a salt bridge that prevents the mixing of the solution.
In Daniel cell flow of current takes place from copper electrode to zinc electrode.
Hence option D is correct.

Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

For the galvanic cell, $Cu|Cu^{2+}||Ag^+|Ag$. Which of the following observations is not correct?

  1. Cu acts as anode and Ag acts as cathode

  2. Ag electrode loses mass and Cu electrode gains mass
  3. Reaction at anode, $Cu\rightarrow Cu^{2+}+2e^-$

  4. Copper is more reactive than silver
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Redox reactions are reactions in which both oxidation and reduction takes place together. In a Galvanic cell oxidation occurs at anode and reduction occurs at cathode. Here $Cu$ is an anode and $Ag$ is a cathode. As in Oxidation, the electrons get released to form aqueous ions in the solution and hence reduces the mass of the electrode as the electrons are released from the electrode. Here,$Cu$ electrode loses mass. Similarly,the released electrons are gained by silver ions i.e $Ag^ { +} $ and the electrode mass of $Ag$ gets increased. Hence, B is the wrong option.

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

One gm metal $M^{3+}$ was discharged by the passage of $1.81\times 10^{23}$ electrons. What is the atomic mass of metal?

  1. $8g/mol$
  2. $9g/mol$
  3. $10g/mol$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Applying faraday's first law-
$ \dfrac{Q}{F} = \dfrac{Wt}{Mwt}\times V.f.$
$ \dfrac{ne}{F}= \dfrac{1}{Mwt}\times 3$
$  Mwt         = \dfrac{1\times F\times 3}{ne}$
$  Mwt         = 10\dfrac{gm}{mol}$
Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Number of electrons involved in the electrodeposition of 63.5 g of Cu from solution of $ CuSO _{4} $ is:

  1. $ 6.022\times 10^{23} $
  2. $ 3.011\times 10^{23} $
  3. $ 12.044\times 10^{23} $
  4. $ 6.022\times 10^{23} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$ Cu^{2+}+2e^-\rightarrow Cu $ 

1 mole Cu = 63.5 g
 
For deposition of 1 mole of Cu, 2 moles of $e^-$ is required.

$ \therefore$ Number of electrons $ =12.044\times 10^{23} $      
                                    
Hence, option C is correct.