IMO Practice Test: Electrochemistry | Class 12 Chemistry

Question 1 of 12

For a Daniell cell, the EMF increases when:

$[\text{Cu}^{2+}]$ is decreased

$[\text{Zn}^{2+}]$ is decreased

both ions are increased equally

temperature is lowered to 0 K

Explanation: $E = E^0 - \frac{0.059}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$; lowering $[\text{Zn}^{2+}]$ makes $\log Q$ more negative, raising $E$.

Question 2 of 12

At equilibrium, the EMF of a galvanic cell becomes:

maximum

zero

negative infinity

equal to $E^0$

Explanation: At equilibrium $Q = K_c$ and the cell does no more work, so $E_{cell} = 0$.

Question 3 of 12

If $E^0_{cell} = 0.295\,\text{V}$ for a 2-electron cell, the value of $\log K_c$ at 298 K is about:

5

10

20

2

Explanation: $\log K_c = \frac{nE^0}{0.059} = \frac{2 \times 0.295}{0.059} = 10$.

Question 4 of 12

The same charge is passed through three cells in series containing $\text{Ag}^+$, $\text{Cu}^{2+}$ and $\text{Al}^{3+}$. The moles deposited are in the ratio:

1 : 1 : 1

6 : 3 : 2

1 : 2 : 3

3 : 2 : 1

Explanation: Moles $\propto 1/n$: Ag $1/1$, Cu $1/2$, Al $1/3$ → multiply by 6 → 6 : 3 : 2.

Question 5 of 12

A metal of equivalent mass 12 is deposited by 1930 C of charge. The mass deposited is:

$0.12\,\text{g}$

$0.24\,\text{g}$

$1.2\,\text{g}$

$2.4\,\text{g}$

Explanation: $m = \frac{E \times Q}{F} = \frac{12 \times 1930}{96500} = 0.24\,\text{g}$.

Question 6 of 12

For a strong electrolyte $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$. A plot of $\Lambda_m$ versus $\sqrt{C}$ has intercept:

$A$

zero

$\Lambda_m^0$

$\sqrt{C}$

Explanation: At $C \to 0$, $\sqrt{C} \to 0$, so the intercept equals $\Lambda_m^0$.

Question 7 of 12

Two H–O fuel cells in series each give $1.23\,\text{V}$. The combined EMF is:

$1.23\,\text{V}$

$0.615\,\text{V}$

$2.46\,\text{V}$

$0\,\text{V}$

Explanation: EMFs add in series: $2 \times 1.23 = 2.46\,\text{V}$.

Question 8 of 12

Galvanised iron resists rusting even when the zinc layer is scratched because:

zinc is less reactive than iron

zinc still acts as a sacrificial anode

iron becomes the anode

the scratch reseals itself

Explanation: Zinc is more reactive, so it keeps corroding preferentially, protecting the exposed iron cathodically.

Question 9 of 12

The conductivity of $0.001\,\text{M}$ acetic acid is $4.95 \times 10^{-5}\,\text{S cm}^{-1}$. Its molar conductivity is:

$49.5\,\text{S cm}^2\,\text{mol}^{-1}$

$4.95\,\text{S cm}^2\,\text{mol}^{-1}$

$495\,\text{S cm}^2\,\text{mol}^{-1}$

$0.495\,\text{S cm}^2\,\text{mol}^{-1}$

Explanation: $\Lambda_m = \frac{4.95 \times 10^{-5} \times 1000}{0.001} = 49.5\,\text{S cm}^2\,\text{mol}^{-1}$.

Question 10 of 12

Which combination gives the highest cell EMF? ($E^0$: $\text{Mg}^{2+}/\text{Mg} = -2.37$, $\text{Zn}^{2+}/\text{Zn} = -0.76$, $\text{Ag}^+/\text{Ag} = +0.80$)

Zn | Ag cell

Mg | Zn cell

Mg | Ag cell

Zn | Mg cell

Explanation: Largest $E^0_{cell} = E_{cathode} - E_{anode} = 0.80 - (-2.37) = 3.17\,\text{V}$ for the Mg | Ag cell.

Question 11 of 12

During the discharge of a lead storage battery, the density of the sulphuric acid:

increases

decreases

stays constant

first rises then falls

Explanation: $\text{H}_2\text{SO}_4$ is consumed and water is formed, so the acid is diluted and its density falls.

Question 12 of 12

The quantity of electricity needed to reduce 1 mol of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ is:

$1\,F$

$3\,F$

$5\,F$

$7\,F$

Explanation: Mn goes from +7 to +2, a gain of 5 electrons, so $5\,F$ are required.

IMO Practice Test — Electrochemistry