A galvanic cell is constructed using the redox reaction, $\frac{1}{2}{{\rm{H}}_2}\left( g \right) + {\rm{AgCl}}\left( s \right)\leftrightharpoons{{\rm{H}}^ + }\left( {aq} \right) + {\rm{C}}{{\rm{l}}^ - }\left( {aq} \right) + {\rm{Ag}}\left( s \right)$ It is represented as

${\rm{Pt}}\left| {{{\rm{H}}_2}\left( {\rm{g}} \right)|{\rm{HCl {\kern 1pt} {\kern 1pt}solution}}} \right|\left| {{\rm{AgCl}}\left( {\rm{s}} \right)} \right|{\rm{Ag}}$

${\rm{Pt}}\left| {{{\rm{H}}_2}\left( {\rm{g}} \right)\left| {{\rm{HCl {\kern 1pt} {\kern 1pt}solution}}} \right|} \right|{\rm{AgN}}{{\rm{O}}_3}{\kern 1pt} {\kern 1pt}{\rm{solution}}|{\rm{Ag}}$

${\rm{Ag}}\left| {{\rm{AgCl}}\left( {\rm{s}} \right)} \right|{\rm{KCl {\kern 1pt} {\kern 1pt}solution}}\parallel {\rm{HCl {\kern 1pt} {\kern 1pt}solution}}\left| {{{\rm{H}}_2}\left( {\rm{g}} \right)} \right|{\rm{Pt}}$

${\rm{Pt}}|{{\rm{H}}_2}\left( {\rm{g}} \right)\left| {{\rm{KCl {\kern 1pt} {\kern 1pt}solution}}} \right|\left| {{\rm{AgCl}}\left( {\rm{s}} \right)} \right|{\rm{Ag}}$

When 1 F of electricity is passed through acidulated water, ${{\rm{O}}_2}$ evolved is

$5.6{\kern 1pt} {\kern 1pt} {\rm{d}}{\kern 1pt} {{\rm{m}}^3}$

$1.0{\kern 1pt} {\kern 1pt} {\rm{d}}{\kern 1pt} {{\rm{m}}^3}$

$11.2{\kern 1pt} {\kern 1pt} {\rm{d}}{\kern 1pt} {{\rm{m}}^3}$

$22.4{\kern 1pt} {\kern 1pt} {\rm{d}}{\kern 1pt} {{\rm{m}}^3}$

${E^ \circ }$ values of $M{g^{(2 + )}}$ / Mg is -2.37 V, of ${\rm{Z}}{{\rm{n}}^{2 + }}$ / Zn is -0.76 V and ${\rm{F}}{{\rm{e}}^{2 + }}$ / Fe is -0.44 V. Which of the statements is correct?

Zn will reduce ${\rm{F}}{{\rm{e}}^{2 + }}$

The cell reaction of the galvanic cell ${\rm{Cu}}\left( s \right)\left| {{\rm{C}}{{\rm{u}}^{2 + }}\left( {aq} \right)} \right|\left| {{\rm{H}}{{\rm{g}}^{2 + }}\left( {aq} \right)} \right|{\rm{Hg}}\left( l \right)$ is

${\rm{Cu}} + {\rm{H}}{{\rm{g}}^{2 + }} \to {\rm{C}}{{\rm{u}}^{2 + }} + {\rm{Hg}}$

${\rm{Hg}} + {\rm{C}}{{\rm{u}}^{2 + }} \to {\rm{H}}{{\rm{g}}^{2 + }} + {\rm{Cu}}$

${\rm{Hg}} + {\rm{C}}{{\rm{u}}^{2 + }} \to {\rm{C}}{{\rm{u}}^ + } + {\rm{H}}{{\rm{g}}^ + }$

${\rm{Cu}} + {\rm{Hg}} \to {\rm{CuHg}}$

The conductivity of $N/50$ solution of ${\rm{KCl}}$ in a cell at 250C is $0.002765{\kern 1pt} {\kern 1pt} {\rm{mho {\kern 1pt}{\kern 1pt} c}}{{\rm{m}}^{ - 1}}$ . If the resistance of a cell containing this solution is 400 ohm, the cell constant is :

$1.106{\kern 1pt} {\kern 1pt} {\rm{c}}{{\rm{m}}^{ - 1}}$

$1.106{\kern 1pt} {\kern 1pt} {\rm {cm}}$

$1{\kern 1pt} {\kern 1pt} {\rm{cm}}$

$1{\kern 1pt} {\kern 1pt} {\rm{c}}{{\rm{m}}^{ - 1}}$

Which substance is obtained in the solution on electrolysis of aqueous ${\rm{CuS}}{{\rm{O}}_4}$ solution using graphite electrodes?

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$

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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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121 Medium

121 of 145

Application +4 | -1

Calculate the emf of the cell

\mathrm{Cu}\left( s \right)\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right)} \right|\left| {\mathrm{A}{\mathrm{g}^ + }\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right)

Given,

E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{cu}}^ \circ = + 0.34V,E_{\mathrm{A}{\mathrm{g}^ + }/\mathrm{Ag}}^ \circ = 0.80V

View answer and explanation

Correct option: A

+0.46 V

Option A (+0.46 V) is correct because it is the calculated EMF of the cell using the given standard reduction potentials.

122 Medium

122 of 145

Application +4 | -1

The weight of silver (eq. wt. = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is :

View answer and explanation

Correct option: A

54 g

Option A (54 g) is correct because the mass of silver displaced is calculated using the equivalent weights and volumes of hydrogen.

123 Medium

123 of 145

Analyzing +4 | -1

In which of the following pairs, the constants/ quantities are not mathematically related to each other?

View answer and explanation

Correct option: D

Rate constant and standard cell potential

Option D (Rate constant and standard cell potential) is correct because there is no direct mathematical relationship between the rate constant of a reaction and the standard cell potential.

124 Medium

124 of 145

Understanding +4 | -1

A galvanic cell is constructed using the redox reaction,

\frac{1}{2}{\mathrm{H}_2}\left( g \right) + \mathrm{AgCl}\left( s \right)\leftrightharpoons{\mathrm{H}^ + }\left( {aq} \right) + \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right) + \mathrm{Ag}\left( s \right)

It is represented as

View answer and explanation

Correct option: D

\mathrm{Pt}\left| {{\mathrm{H}_2}\left( \mathrm{g} \right)|{\rm{HCl {\kern 1pt} {\kern 1pt}solution}}} \right|\left| {\mathrm{AgCl}\left( \mathrm{s} \right)} \right|\mathrm{Ag}

Option D (

\mathrm{Pt}\left| {{\mathrm{H}_2}\left( \mathrm{g} \right)|{\rm{HCl {\kern 1pt} {\kern 1pt}solution}}} \right|\left| {\mathrm{AgCl}\left( \mathrm{s} \right)} \right|\mathrm{Ag}

) is correct because it correctly represents the cell notation for the given redox reaction.

125 Medium

125 of 145

Application +4 | -1

Calculate the pH of

C{H_3}COOH

solution, if equivalent conductance of

1M\,\,C{H_3}COOH

10\,oh{m^{ - 1}}c{m^2}e{q^{ - 1}}

and at infinite solution is

200\,oh{m^{ - 1}}c{m^2}e{q^{ - 1}}

(Given : log 5 = 0.7)

View answer and explanation

Correct option: B

1.3

Option B is correct because the pH is calculated using the degree of dissociation (α) and the concentration of the acid, resulting in a pH of approximately 1.3.

126 Medium

126 of 145

Application +4 | -1

When 1 F of electricity is passed through acidulated water,

{\mathrm{O}_2}

evolved is

View answer and explanation

Correct option: B

5.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}

Option B (

5.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}

) is correct because 1 F of electricity passed through acidulated water evolves 0.25 moles of

O_2

, which occupies 5.6 L (or

5.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}

) at STP.

127 Medium

127 of 145

Analyzing +4 | -1

250 mL of a waste solution obtained from the workshop of a goldsmith contains

0.1\,M\,AgN{O_3}

and

0.1\,M\,AgCl.

The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electro deposited will be:

\left( {E_{A{g^ + }/Ag}^0 = 0.80\,V,E_{A{u^ + }/Au}^0\, = 1.69\,V} \right)

View answer and explanation

Correct option: A

Only gold

Option A (Only gold) is correct because gold has a higher reduction potential than silver, meaning silver will deposit first, and the applied voltage is sufficient to deposit gold.

128 Medium

128 of 145

Understanding +4 | -1

When an electric current is passed through an aqueous solution of sodium chloride :

View answer and explanation

Correct option: D

Its pH progressively increases

Option D (Its pH progressively increases) is correct because during the electrolysis of aqueous NaCl, hydrogen gas is produced at the cathode, leading to an increase in hydroxide ion concentration and thus an increase in pH.

129 Medium

129 of 145

Analyzing +4 | -1

{E^ \circ }

values of

M{g^{(2 + )}}

/ Mg is -2.37 V, of

\mathrm{Z}{\mathrm{n}^{2 + }}

/ Zn is -0.76 V and

\mathrm{F}{\mathrm{e}^{2 + }}

/ Fe is -0.44 V.
Which of the statements is correct?

View answer and explanation

Correct option: A

Zn will reduce

\mathrm{F}{\mathrm{e}^{2 + }}

Option A (Zn will reduce Fe2+) is correct because Zn has a more negative reduction potential than Fe, making it a stronger reducing agent.

130 Medium

130 of 145

Understanding +4 | -1

The cell reaction of the galvanic cell

\mathrm{Cu}\left( s \right)\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right)} \right|\left| {\mathrm{H}{\mathrm{g}^{2 + }}\left( {aq} \right)} \right|\mathrm{Hg}\left( l \right)

View answer and explanation

Correct option: D

\mathrm{Cu} + \mathrm{H}{\mathrm{g}^{2 + }} \to \mathrm{C}{\mathrm{u}^{2 + }} + \mathrm{Hg}

Option D (

\mathrm{Cu} + \mathrm{H}{\mathrm{g}^{2 + }} \to \mathrm{C}{\mathrm{u}^{2 + }} + \mathrm{Hg}

) is correct because copper is oxidized to

Cu^{2+}

and

Hg^{2+}

is reduced to Hg.

131 Medium

131 of 145

Application +4 | -1

Assume that during electrolysis of

AgN{O_3}

, only

{H_2}O

is electrolysed and

{O_2}

is formed as:

2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }

{O_2}

formed at NTP due to passage of 2A of current for 965 s is

View answer and explanation

Correct option: A

0.112 L

Option A is correct. The volume of oxygen formed is calculated using Faraday's laws, considering the current, time, and the stoichiometry of the electrolysis reaction of water.

132 Medium

132 of 145

Understanding +4 | -1

A cell necessarily does not contain :

View answer and explanation

Correct option: B

A cathode

Option D (A porous diaphragm) is correct because while anodes, cathodes, and electrolytes/fuels are essential components of a cell, a porous diaphragm is not always necessary; some cells function without it.

133 Medium

133 of 145

Remembering +4 | -1

Which of the following metal has the highest conductivity at room temperature ?

View answer and explanation

Correct option: C

Option C (Ag) is correct because silver (Ag) has the highest electrical conductivity at room temperature among the given options.

134 Medium

134 of 145

Understanding +4 | -1

Normal

\mathrm{Al} - \mathrm{AlC}{\mathrm{l}_3}

coupled with normal hydrogen electrode gives an e.m.f. of 1.66 V. The standard oxidation electrode potential of aluminium is :

View answer and explanation

Correct option: B

+ 1.66{\kern 1pt} {\kern 1pt} \mathrm{V}

Option B (

+ 1.66{\kern 1pt} {\kern 1pt} \mathrm{V}

) is correct because the standard oxidation potential is the negative of the standard reduction potential, and since the emf is given as 1.66 V, the oxidation potential of aluminum is +1.66 V.

135 Medium

135 of 145

Remembering +4 | -1

According to Faraday’s first law of electrolysis mass of substances liberated is equal to

View answer and explanation

Correct option: B

Option B is correct because, according to Faraday's first law of electrolysis, the mass of a substance liberated at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte, represented by the equation m = eQ, where e is the electrochemical equivalent and Q is the charge.

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136 Medium

136 of 145

Application +4 | -1

The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant

0.66\mathrm{c}{\mathrm{m}^{ - 1}}

. The equivalent conductivity of solution is :

View answer and explanation

Correct option: A

314.28{\kern 1pt}{\kern 1pt}{\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}

Option A (

314.28{\kern 1pt}{\kern 1pt}{\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}

) is correct because it is the result of correctly calculating the equivalent conductivity using the given resistance, cell constant, and normality.

137 Medium

137 of 145

Analyzing +4 | -1

Given that

\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\circ}=-0.36 \mathrm{~V}

and

\mathrm{E}_{\mathrm{Fe}^{2+}/\mathrm{Fe}}^{\circ}=-0.439 \mathrm{~V}

respectively. Therefore value of

\mathrm{E}_{\mathrm{F}^{3+}, \mathrm{Fe}^{2+} /\mathrm{Pt}}^{\mathrm{o}}

is:

View answer and explanation

Correct option: D

[3(-0.036)-2(-

0.439)] V

Option D (

[3(-0.036)-2(-

0.439)] V) is correct because it correctly applies the formula for calculating the standard reduction potential of the overall reaction using the given standard reduction potentials and the number of electrons involved.

138 Medium

138 of 145

Application +4 | -1

For the electrochemical cell,

M\left| {{M^ + }||{X^ - }} \right|X,

{E^ \circ }({M^ + }|M) = 0.44V,{E^ \circ } = (X|{X^ - }) = 0.33V

. From this data one can deduce that

View answer and explanation

Correct option: B

{M^ + } + {X^ - } \to M + X

is the spontaneous reaction

Option C (

M + X \to {M^ + } + {X^ - }

is the spontaneous reaction) is correct because the cell potential is positive, indicating a spontaneous reaction in the direction written.

139 Medium

139 of 145

Analyzing +4 | -1

During electrolysis of sulphuric acid, which of the following processes is possible at anode?

A.

2 {H}_2 {O}({l}) \longrightarrow {O}_2({~g})+4 {H}^{+}({aq})+4 {e}^{-} \stackrel{\ominus}{{E}_{\text {cell }}}=+1.23 {~V}

2 {SO}_4^{2-}({aq}) \longrightarrow {S}_2 {O}_8^{2-}({aq})+2 {e}^{-} \stackrel{\ominus}{{E}_{\text {cell }}}= 1.96 {~V}

Choose the correct option based on following statements.

(i) Process A is preferred at higher concentration of sulphuric acid.

(ii) Process B is preferred at higher concentration of sulphuric acid.

(iii) Process A is preferred for dilute sulphuric acid.

(iv) Process B is preferred for dilute sulphuric acid.

(v) Both A and B are equally possible at higher concentration.

View answer and explanation

Correct option: B

(iii) and (ii)

Option B ((iii) and (ii)) is correct because at higher concentrations of sulphuric acid, process B is preferred, and for dilute sulphuric acid, process A is preferred.

140 Medium

140 of 145

Analyzing +4 | -1

Assertion: Conductivity of metals decreases with increase of temperature whereas that of electrolytic solution increases.

Reason:Electrons in metals are very tightly held by the nucleus and are not free to move.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

Option C is correct because the assertion is true (metals' conductivity decreases with temperature, while electrolytic solutions' increases), but the reason is false (electrons in metals are free to move, not tightly held).

141 Medium

141 of 145

Application +4 | -1

For the reduction of silver ions with copper metal, the standard cell potential is 0.46 V at

{25^ \circ }C

. The value of standard Gibbs energy

\Delta {G^ \circ }

will be :

View answer and explanation

Correct option: A

- 89.0{\kern 1pt} {\kern 1pt} \mathrm{kJ}

Option A (

- 89.0{\kern 1pt} {\kern 1pt} \mathrm{kJ}

) is correct because it is the result of correctly applying the formula

\Delta G^\circ = -nFE^\circ

with the correct number of electrons transferred.

142 Medium

142 of 145

Application +4 | -1

{10^{ - 2}}

g atom of Ag can be oxidised to

\mathrm{A}{\mathrm{g}^ + }

during the electrolysis of

\mathrm{AgN}{\mathrm{O}_3}

solution using silver electrode by :

View answer and explanation

Correct option: A

965 coulomb

Option A (965 coulomb) is correct because the charge required is directly proportional to the number of gram atoms oxidized, based on Faraday's laws.

143 Medium

143 of 145

Application +4 | -1

The conductivity of

N/50

solution of

\mathrm{KCl}

in a cell at 250C is

0.002765{\kern 1pt} {\kern 1pt} {\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^{ - 1}}

. If the resistance of a cell containing this solution is 400 ohm, the cell constant is :

View answer and explanation

Correct option: B

1.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}

Option B (

1.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}

) is correct because the cell constant is calculated by multiplying conductivity and resistance.

144 Medium

144 of 145

Understanding +4 | -1

Which substance is obtained in the solution on electrolysis of aqueous

\mathrm{CuS}{\mathrm{O}_4}

solution using graphite electrodes?

View answer and explanation

Correct option: A

{\mathrm{H}_2}\mathrm{O}

Option B (

H_2SO_4

) is correct because during the electrolysis of aqueous CuSO4 using graphite electrodes, water is oxidized at the anode to produce oxygen gas and sulfuric acid remains in the solution.

145 Medium

145 of 145

Remembering +4 | -1

Which of the following electrolytes is used in lead storage battery?

View answer and explanation

Correct option: C

38 % solution of

{H_2}S{O_4}

Option C (38 % solution of

{H_2}S{O_4}

) is correct because dilute sulfuric acid is the electrolyte used in lead storage batteries.

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