Online Test: Electrochemistry | Class 12 Chemistry

Question 1 of 18

In a galvanic cell, the electrode at which oxidation occurs is the:

cathode

anode

salt bridge

electrolyte

Explanation: Oxidation always takes place at the anode (the negative electrode in a galvanic cell).

Question 2 of 18

The standard EMF of the Daniell cell is:

$0.34\,\text{V}$

$0.76\,\text{V}$

$1.10\,\text{V}$

$1.50\,\text{V}$

Explanation: $E^0_{cell} = 0.34 - (-0.76) = 1.10\,\text{V}$.

Question 3 of 18

The potential of the standard hydrogen electrode is taken as:

$+1\,\text{V}$

$-1\,\text{V}$

$0\,\text{V}$

$0.059\,\text{V}$

Explanation: The SHE is the reference electrode, assigned exactly zero volts.

Question 4 of 18

The relation between Gibbs energy and cell EMF is:

$\Delta G = nFE_{cell}$

$\Delta G = -nFE_{cell}$

$\Delta G = -\frac{nE_{cell}}{F}$

$\Delta G = nE_{cell}/F$

Explanation: $\Delta G = -nFE_{cell}$; a positive EMF gives a negative $\Delta G$.

Question 5 of 18

In the Nernst equation at 298 K, the coefficient of $\frac{1}{n}\log Q$ is:

$0.059$

$0.59$

$2.303$

$96500$

Explanation: $E = E^0 - \frac{0.059}{n}\log Q$ at 298 K.

Question 6 of 18

For the cell reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, the number of electrons transferred is:

1

2

3

4

Explanation: Both Zn and Cu change by 2 units of charge, so $n = 2$.

Question 7 of 18

On dilution, the molar conductivity of an electrolyte:

decreases

increases

stays constant

becomes zero

Explanation: Molar conductivity rises on dilution as ions move more freely or dissociate more.

Question 8 of 18

The unit of conductivity ($\kappa$) is:

$\text{S cm}^2\,\text{mol}^{-1}$

$\text{S cm}^{-1}$

$\Omega$

$\text{mol L}^{-1}$

Explanation: Conductivity is measured in $\text{S cm}^{-1}$ (or $\text{S m}^{-1}$).

Question 9 of 18

Molar conductivity $\Lambda_m$ is calculated as:

$\kappa \times C$

$\frac{\kappa \times 1000}{C}$

$\frac{C}{\kappa}$

$\kappa + C$

Explanation: $\Lambda_m = \frac{\kappa \times 1000}{C}$ with $\kappa$ in $\text{S cm}^{-1}$ and $C$ in $\text{mol L}^{-1}$.

Question 10 of 18

Kohlrausch's law of independent migration of ions applies at:

high concentration

infinite dilution

the boiling point

zero temperature

Explanation: At infinite dilution ions move independently, so $\Lambda_m^0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0$.

Question 11 of 18

The degree of dissociation of a weak electrolyte equals:

$\Lambda_m^0/\Lambda_m$

$\Lambda_m/\Lambda_m^0$

$\Lambda_m \times \Lambda_m^0$

$\Lambda_m - \Lambda_m^0$

Explanation: $\alpha = \Lambda_m/\Lambda_m^0$ — observed over limiting molar conductivity.

Question 12 of 18

The charge required to deposit one mole of $\text{Al}$ from $\text{Al}^{3+}$ is:

$96500\,\text{C}$

$2 \times 96500\,\text{C}$

$3 \times 96500\,\text{C}$

$\frac{96500}{3}\,\text{C}$

Explanation: $\text{Al}^{3+} + 3e^-\rightarrow\text{Al}$ needs 3 moles of electrons = $3F$.

Question 13 of 18

According to Faraday's first law, the mass deposited is proportional to:

the voltage

the charge passed

the resistance

the cell volume

Explanation: $m = ZIt$, so $m \propto Q = It$.

Question 14 of 18

Which of the following is a primary cell?

lead storage battery

Ni–Cd cell

dry cell

fuel cell

Explanation: The dry (Leclanché) cell is non-rechargeable; lead storage and Ni–Cd are secondary cells.

Question 15 of 18

The cathode reaction in a hydrogen–oxygen fuel cell (alkaline) is:

$\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$

$\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$

$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

Explanation: Oxygen is reduced at the cathode: $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$.

Question 16 of 18

Rusting of iron is prevented by connecting it to a block of:

copper

silver

magnesium

gold

Explanation: Magnesium, being more reactive, acts as a sacrificial anode (cathodic protection).

Question 17 of 18

A more negative standard reduction potential means the species is a:

stronger oxidising agent

stronger reducing agent

weaker reducing agent

non-electrolyte

Explanation: More negative $E^0$ means a stronger tendency to be oxidised, i.e. a stronger reducing agent.

Question 18 of 18

When the same quantity of charge is passed through solutions of $\text{AgNO}_3$ and $\text{CuSO}_4$, the ratio of moles of Ag to Cu deposited is:

1 : 1

2 : 1

1 : 2

3 : 1

Explanation: $\text{Ag}^+$ needs 1 e⁻ but $\text{Cu}^{2+}$ needs 2 e⁻, so for equal charge twice as many Ag are deposited: 2 : 1.

Online Test — Electrochemistry