Electrochemistry — Page 7 | JEE Chemistry

Q61

Which of the following statements is not correct about rusting of iron?

A Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.

B Dissolved acidic oxides

\mathrm{SO}_2, \mathrm{NO}_2

in water act as catalyst in the process of rusting.

C Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.

D When

\mathrm{pH}

lies above 9 or 10, rusting of iron does not take place.

Correct Answer

Option C

Solution

The statement that is not correct about the rusting of iron is Option C.

Let's examine each option one by one: Option A: Rusting of iron is envisaged as setting up of an electrochemical cell on the surface of iron object.

This statement is correct.

Rusting of iron is an electrochemical process that occurs when iron (Fe) reacts with water (H2O) and oxygen (O2) to form hydrated iron(III) oxide, which we commonly know as rust.

Iron acts as the anode, the part where oxidation occurs as iron loses electrons.

Oxygen, often dissolved in water, acts as the cathode where reduction occurs, completing the electrochemical cell.

Option B: Dissolved acidic oxides such as

\mathrm{SO}_2

and

\mathrm{NO}_2

in water act as catalyst in the process of rusting. This statement is correct. Acidic oxides like

\mathrm{SO}_2

and

\mathrm{NO}_2

can dissolve in water to form acidic solutions which can lead to the acceleration of the rate of rusting as they may lower the pH of water, thereby increasing the availability of hydrogen ions (H+) which may facilitate the electrochemical reactions involved in rusting.

Option C: Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.

This statement is incorrect.

Tin (Sn) is used as a protective coating for iron to prevent it from rusting; however, if the tin coating starts to peel off, it exposes the iron underneath to the atmosphere, which will then be susceptible to rusting.

Once the protective layer is compromised, the iron is at risk, especially if the exposed parts are more anodic than tin.

Tin as a more cathodic metal can even act to accelerate the corrosion of the exposed iron—this is known as a bimetallic corrosion or galvanic corrosion.

Option D: When

\mathrm{pH}

lies above 9 or 10, rusting of iron does not take place.

This statement is generally correct.

Rusting of iron is lessened in alkaline conditions, i.e., when the pH is above 9 or 10 because the increased availability of hydroxide ions (OH-) can reduce the solubility of iron(II) and iron(III) ions, leading to their precipitation as hydroxides rather than participating in the rusting process.

Therefore, the statement from the given options that is not correct about the rusting of iron is Option C, which falsely claims that a peeling tin surface can still prevent rusting of iron.

Q62

Identify the factor from the following that does not affect electrolytic conductance of a solution.

A The nature of solvent used.

B The nature of the electrolyte added.

C The nature of the electrode used.

D Concentration of the electrolyte.

Correct Answer

Option C

Solution

Conductivity of electrolytic cell is affected by concentration of electrolyte, nature of electrolyte and nature of solvent.

Q63

Alkaline oxidative fusion of

\mathrm{MnO}_2

gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are

A

\mathrm{Mn}_2 \mathrm{O}_3

and

\mathrm{MnO}_4^{2-}

B

\mathrm{Mn}_2 \mathrm{O}_7

and

\mathrm{MnO}_4^{-}

C

\mathrm{MnO}_4^{2-}

and

\mathrm{MnO}_4^{-}

D

\mathrm{MnO}_4^{2-}

and

\mathrm{Mn}_2 \mathrm{O}_7

Correct Answer

Option C

Solution

Alkaline oxidative fusion of

\mathrm{MnO}_2

:

2 \mathrm{MnO}_2+4 \mathrm{OH}^{-}+\mathrm{O}_2 \rightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{H}_2 \mathrm{O}

Electrolytic oxidation of

\mathrm{MnO}_4^{2-}

in alkaline medium.

\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^{-}+e^{-}

Q64

Reduction potential of ions are given below:

\begin{array}{ccc} \mathrm{ClO}_4^{-} & \mathrm{IO}_4^{-} & \mathrm{BrO}_4^{-} \\ \mathrm{E}^{\circ}=1.19 \mathrm{~V} & \mathrm{E}^{\circ}=1.65 \mathrm{~V} & \mathrm{E}^{\circ}=1.74 \mathrm{~V} \end{array}

The correct order of their oxidising power is :

A

\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}

B

\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}

C

\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}

D

\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}

Correct Answer

Option D

Solution

Higher the value of $\oplus$ ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant.

Hence, ox.

Power:-

\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}

Q65

One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?

A Metal ion - Metal electrodes

B Oxidation - Reduction electrodes

C Metal - Insoluble Salt - Anion electrodes

D Gas - Ion electrodes

Correct Answer

Option C

Solution

Calomel electrode is metal-insoluble salt – Anion electrode.

Q66

Which of the following electrolyte can be used to obtain

\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8

by the process of electrolysis ?

A Concentrated solution of sulphuric acid

B Dilute solution of sodium sulphate.

C Acidified dilute solution of sodium sulphate.

D Dilute solution of sulphuric acid

Correct Answer

Option A

Solution

$\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ is obtained by electrolysis of concentrated solution of sulphuric acid.

At anode :

2 \mathrm{HSO}_4^{-} \rightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{e}^{-}

Q67

What pressure (bar) of

\mathrm{H}_2

would be required to make emf of hydrogen electrode zero in pure water at

25^{\circ} \mathrm{C}

?

A 0.5

B

10^{-14}

C 1

D

10^{-7}

Correct Answer

Option B

Solution

The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction

\mathrm{H_2(g)} \rightarrow 2\mathrm{H^+(aq)} + 2\mathrm{e^-}

is given by:

E = E^\circ + \frac{RT}{2F} \ln \left( \frac{[\mathrm{H^+}]^2}{P_{\mathrm{H_2}}} \right)

where: E is the electrode potential.

E∘ is the standard electrode potential, which is 0 V for the hydrogen electrode.

R is the gas constant,

8.314 \, \mathrm{J \cdot mol^{-1} \cdot K^{-1}}

. T is the temperature in Kelvin:

25^{\circ} \mathrm{C} = 298 \, \mathrm{K}

. F is the Faraday constant,

96485 \, \mathrm{C \cdot mol^{-1}}

.

[\mathrm{H^+}]

is the concentration of hydrogen ions.

P_{\mathrm{H_2}}

is the pressure of hydrogen gas. In pure water at

25^{\circ} \mathrm{C}

,

[\mathrm{H^+}] = 10^{-7} \, \mathrm{M}

. To make the emf zero, we set E to 0 in the Nernst equation:

0 = 0 + \frac{8.314 \times 298}{2 \times 96485} \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right)

Simplifying the constants:

\frac{8.314 \times 298}{2 \times 96485} \approx 0.0128

Thus, the equation becomes:

0 = 0.0128 \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right)

Since ln term must be zero for this equation to hold true (as 0 divided by any number is still 0), the expression inside the logarithm must equal 1:

\frac{(10^{-7})^2}{P_{\mathrm{H_2}}} = 1

Simplifying this gives:

(10^{-7})^2 = P_{\mathrm{H_2}}

10^{-14} = P_{\mathrm{H_2}}

Therefore, the required pressure of

\mathrm{H_2}

to make the emf of the hydrogen electrode zero in pure water at

25^{\circ} \mathrm{C}

is

10^{-14}

bar. The correct answer is Option B:

10^{-14}

.

Q68

Fuel cell, using hydrogen and oxygen as fuels, A. has been used in spaceship B. has as efficiency of

40 \%

to produce electricity C. uses aluminum as catalysts D. is eco-friendly E. is actually a type of Galvanic cell only Choose the correct answer from the options given below:

A A, D, E only

B A, B, C only

C A, B, D only

D A, B, D, E only

Correct Answer

Option A

Solution

Fuel cells produce electricity with an efficiency of about

70 \%

. Fuel cells are pollution free, thus, eco-friendly. Fuel cells are type of Galvanic cells only.

Q69

For a strong electrolyte, a plot of molar conductivity against (concentration)

{ }^{1 / 2}

is a straight line, with a negative slope, the correct unit for the slope is

A

\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}

B

\mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{-1 / 2}

C

\mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-1} \mathrm{~L}^{1 / 2}

D

\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{1 / 2}

Correct Answer

Option D

Solution

The molar conductivity ( $\Lambda_m$ ) of a strong electrolyte depends on the concentration ( $c$ ) according to Kohlrausch's law, which can be mathematically expressed as $\Lambda_m = \Lambda_m^0 - k\sqrt{c}$ , where $\Lambda_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant.

The graph of molar conductivity ( $\Lambda_m$ ) against the square root of concentration ( $c^{1/2}$ ) is a straight line with a negative slope.

The unit of molar conductivity ( $\Lambda_m$ ) is Siemens meter squared per mole ( $\text{S cm}^2 \text{mol}^{-1}$ ).

Concentration ( $c$ ) has the unit moles per liter ( $\text{mol L}^{-1}$ ), and thus its square root ( $c^{1/2}$ ) has the unit $\text{mol}^{1/2} \text{L}^{-1/2}$ .

The slope of the plot of $\Lambda_m$ against $c^{1/2}$ is derived from the expression $k$ in $k\sqrt{c}$ .

Therefore, to find the correct unit of the slope, we look at the division of the unit of $\Lambda_m$ by the unit of $c^{1/2}$ : $\dfrac{\text{S cm}^2 \text{mol}^{-1}}{\text{mol}^{1/2} \text{L}^{-1/2}} = \text{S cm}^2 \text{mol}^{-1-1/2} \text{L}^{1/2} = \text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$ Thus, the correct unit for the slope of a plot of molar conductivity against the square root of concentration for a strong electrolyte is $\text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$ , which matches with Option D.

Q70

The emf of cell

\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}

is

0.83 \mathrm{~V}

at

298 \mathrm{~K}

. It could be increased by :

A increasing concentration of

\mathrm{Tl}^{+}

ions

B increasing concentration of

\mathrm{Cu}^{2+}

ions

C increasing concentration of both

\mathrm{Tl}^{+}

and

\mathrm{Cu}^{2+}

ions

D decreasing concentration of both

\mathrm{Tl}^{+}

and

\mathrm{Cu}^{2+}

ions

Correct Answer

Option B

Solution

To determine how the emf of the cell,

\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}

, can be increased, we can use the Nernst equation. The Nernst equation for this electrochemical cell is given by:

E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \left( \frac{[\mathrm{Tl}^{+}]}{[\mathrm{Cu}^{2+}]} \right)

where:

E_{\text{cell}}

is the emf of the cell.

E_{\text{cell}}^\circ

is the standard emf of the cell.

n

is the number of moles of electrons transferred in the cell reaction; here,

n = 2

.

[\mathrm{Tl}^{+}]

is the concentration of thallium ions.

[\mathrm{Cu}^{2+}]

is the concentration of copper ions.

Given that the emf of the cell can be expressed in terms of the concentration of the ions involved, we can see that increasing the concentration of

[\mathrm{Cu}^{2+}]

or decreasing the concentration of

[\mathrm{Tl}^{+}]

will affect the logarithmic term in the Nernst equation:

E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{2} \log \left( \frac{0.001}{0.01} \right)

To increase the emf (

E_{\text{cell}}

) of the cell, it is beneficial to have a less negative (or more positive) correction term.

This can be achieved by: Increasing the concentration of

[\mathrm{Cu}^{2+}]

ions: This makes the term inside the logarithm smaller (since we are dividing by a larger number), which in turn makes the logarithmic term less negative.

Therefore, the correct answer is: Option B: increasing concentration of

\mathrm{Cu}^{2+}

ions