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 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 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.
Multiple choice power transmission household electricity household circuits electricity and magnetism physics

A battery is charged at a potential of 15V for 8 hours when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 hours. The mean terminal voltage during discharges is 14 V. The "Watt hour" efficiency of the battery is :-

  1. 80%

  2. 90%

  3. 87.5%

  4. 82.5%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

% Watt hour efficiency = $\frac{E _{out}} {E _{in}} \times 100$
= $\frac{\left ( 14 \right )\left ( 5 \right )\left ( 15 \right )} {\left ( 15 \right )\left ( 10 \right )\left ( 8 \right )} \times 100 = 87.5$%

Multiple choice physics magnetic effects of electric current direct current types of current movement of charge

Which of the following is wrong?

  1. It is not possible to have electrolysis by a.c.

  2. Ordinary batteries cannot be changed by a.c.

  3. In a circuit containing L, C and R in series across a steady source, poer is dissipated in R, at steady state

  4. For current dirven from a steady source by an ohmic conductor, mean value and r.m.s. value of current will be same
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry oxidation- reduction reactions redox reactions and electrode processes redox and electrode processes applications of redox reaction

The negative Zn pole of Daniel cell, sending a constant current through a circuit, decreases in mass by 0.13 g in 30 minutes. If the chemical equivalent of Zn and Cu and 32.5 and 31.5 respectively, the increase in the mass of the positive Cu pole in this time is : 

  1. 0.180 g

  2. 0.141 g

  3. 0.126 g

  4. 0.242 g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Faraday's laws of electrolysis, the mass deposited is proportional to the chemical equivalent. The mass of Cu deposited = (Mass of Zn lost * Equivalent weight of Cu) / Equivalent weight of Zn = (0.13 * 31.5) / 32.5 = 0.126 g.

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

The passage of electricity in the Daniell cell when $Zn$ and $Cu$ electrodes are connected is from:

  1. $Cu$ to $Zn$ in the cell
  2. $Cu$ to $Zn$ outside the cell
  3. $Zn$ to $Cu$ outside the cell
  4. $Zn$ to $Cu$ in the cell
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Daniell cell, Zn acts an anode and Cu acts as cathode, electron flow takes place from anode to cathode and hence, electricity flows from Cu to Zn outside the cell.

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

For the Daniel cell involving the cell reaction${ Zn } _{ (s) }+{ { Cu }^{ +2 } } _{ (aq) }\rightleftharpoons { { { Zn }^{ +2 } } } _{ (aq) }+{ Cu } _{ (s) }$ the standard free energies of formation of  ${ Zn } _{ (s) }$, ${ Cu } _{ (s) }$, ${ { Cu }^{ +2 } } _{ (aq) }$ and ${ { { Zn }^{ +2 } } } _{ (aq) }$ are 0, 0, 64.4 KJ/Mole and -154.0 KJ/Mole, respectively. Calculate the standard EMF of the cell?

  1. $2.13 Volts$
  2. $1.13 Volts$
  3. $2.26 Volts$
  4. $3.42 Volts$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\because \quad \Delta { G }^{ 0 }=\left( { G } _{ { Zn }^{ 2+ } }^{ 0 }-{ G } _{ { Cu }^{ 2+ } }^{ 0 } \right) $ $= (-154-64.4) KJ/Mole$

$\therefore(-2\times{E}^{0}\times F)$ $=(-218.4\times {10}^3)$
$\therefore E^0=1.1316\quad volt$

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.