Electrochemistry — DPP 1

20 NEET questions from Electrochemistry, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

Attempt this test free 20 questions · 30 minutes · +80 marks
1
Which of the following is the strongest reducing agent?
  1. A Li
  2. B Mg
  3. C Al
  4. D Zn
Show answer & explanation

Correct answer: A

The reducing power of an element is determined by its ability to lose electrons.
$\text { Reducing agent } \propto \frac{1}{\text { SRP Value }}$
Lithium has lowest SRP value. Hence, it is the strongest reducing agent.
Therefore, option (1) is the correct answer.
2
Calculate $\mathrm { K } _ { \mathrm { a } }$ of acetic acid if its 0.05 M solution has molar conductivity of $7.814 \times 10 ^ { - 4 } \Omega ^ { - 1 } \mathrm {~m} ^ { 2 } \mathrm {~mol} ^ { - 1 }$ at $25 ^ { \circ } \mathrm { C }$. Given: $\Lambda _ { \mathrm { m } } ^ { \circ }$ for $\mathrm { CH } _ { 3 } \mathrm { COOH } = 3.907 \times 10 ^ { - 2 } \Omega ^ { - 1 } \mathrm { m } ^ { 2 } \mathrm {~mol} ^ { - 1 }$.
  1. A $2 \times 10 ^ { - 5 }$
  2. B $1.8 \times 10 ^ { - 5 }$
  3. C $4 \times 10 ^ { - 4 }$
  4. D 0.02
Show answer

Correct answer: A

3
For the given cell.
$\mathrm{Cd}(\mathrm{~s})\left|\mathrm{Cd}^{2+}(0.1 \mathrm{M}) \| \mathrm{AgNO}_{3}(0.2 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$
The emf of the cell at 298 K is:
Given: $\mathrm{E}^{\circ}{ }_{\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)}=-0.40 \mathrm{V}$ and $\mathrm{E}^{\circ}{ }_{\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)}=0.80 \mathrm{V}$
  1. A 2.96 V
  2. B 1.18 V
  3. C 1.08 V
  4. D 2.08 V
Show answer & explanation

Correct answer: B

Anode : $\mathrm{Cd}(\mathrm{s}) \rightarrow \mathrm{Cd}^{2+}+2 e^{-}$
Cathode : $2 \mathrm{Ag}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}(\mathrm{s})$
Cell reaction : $\mathrm{Cd}(\mathrm{s})+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cd}^{2+}+2 \mathrm{Ag}$

$\mathrm{E}^{\circ}{ }_{\text {Cell }} =\mathrm{E}^{\circ}{ }_{\text {(cathode) }}-\mathrm{E}_{\text {(anode) }}^{\circ} =0.80-(-0.40)=1.2 \mathrm{~V}$

$\mathrm{E}_{\text {cell }} =\mathrm{E}^{\circ}{ }_{\text {Cell }}-\frac{0.059}{\mathrm{n}} \log \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} =1.2-\frac{0.059}{2} \log \frac{0.1}{(0.2)^{2}}$
$=1.2-0.012=1.18 \mathrm{~V}$
Therefore, option (2) is the correct answer.
4
For the given reaction, the value of equilibrium constant is:
$\mathrm{Zn}(\mathrm{~s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$
  1. A $10^{34}$
  2. B $10^{10}$
  3. C $10^{19}$
  4. D $10^{37}$
Show answer & explanation

Correct answer: D

$\begin{array}{l}\mathrm{E}_{\text {Cell }}^{\circ}=\frac{0.059}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}} \\ \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{n} \mathrm{E}^{\circ}}{0.059}\end{array}$
or $\mathrm{K}_{\mathrm{eq}}$ = $10^{\left(\frac{\mathrm{nE}}{0.059}\right)}$
$=10^{\left(\frac{2 \times 1.1}{0.059}\right)}$
$ \approx 10^{37}$
Therefore, option (4) is the correct answer.
5
For an electrochemical cell $\mathrm { Ag } \left| \mathrm { Ag } ^ { + } ( 0.1 \mathrm { M } ) \| \mathrm { Ag } ^ { + } \right| \mathrm { Ag } ( 1 \mathrm { M } )$, which of the following electrolyte cannot be used in salt bridge?
  1. A $\mathrm { NaNO } _ { 3 }$
  2. B KCl
  3. C $\mathrm { KNO } _ { 3 }$
  4. D Both (B) and (C)
Show answer

Correct answer: B

6
A cell from the following which converts electrical energy into chemical energy is called:
  1. A dry cell.
  2. B galvanic cell.
  3. C electrolytic cell.
  4. D None of these.
Show answer

Correct answer: B

7
The solution of $\mathrm{AgNO}_{3}$ in which Ag rod is dipped is increased to $\frac{1}{10}$ times, the reduction electrode potential will:
  1. A decrease by 0.059 V
  2. B decrease by 0.03 V
  3. C increase by 0.059 V
  4. D increase by 0.03 V
Show answer & explanation

Correct answer: C

$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{~s}) \text { Initially, } \mathrm{E}_{\text {cell(Initial) }}=\mathrm{E}_{\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)}^{\circ}-\frac{0.059}{1} \log \frac{1}{\left[\mathrm{Ag}^{+}\right]_{\text {initial }}}$
After dilution,
${\left[\mathrm{Ag}^{+}\right]_{\text {final }}=10\left[\mathrm{Ag}^{+}\right]_{\text {initial }}}$

$\mathrm{E}_{\text {cell(final) }}=\mathrm{E}_{\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)}-\frac{0.059}{1} \log \frac{1}{\left[\mathrm{Ag}^{+}\right]_{\text {initial }}}$

$\mathrm{E}_{\text {cell(final) }}-\mathrm{E}_{\text {cell(initial) }}=0.059 \log \frac{\left[\mathrm{Ag}^{+}\right]_{\text {final }}}{\left[\mathrm{Ag}^{+}\right]_{\text {initial }}}$

$=0.059 \log \left(\frac{10}{1}\right)=+0.059 \mathrm{~V}$
Reduction potential increases by 0.059 V.
Therefore, option (3) is the correct answer.
8
The standard emf of the cell, $\mathrm { Ni } \mid \mathrm { Ni } ^ { 2 + } ( 1.0 \mathrm { M } ) \| \mathrm { Ag } ^ { + } ( 1.0 \mathrm { M } ) \mid \mathrm { Ag } \left[ \mathrm { E } ^ { \circ } \right.$ for $\mathrm { Ni } ^ { 2 + } / \mathrm { Ni } = - 0.25$ volt, $\mathrm { E } ^ { \circ }$ for $\mathrm { Ag } ^ { + } / \mathrm { Ag } = 0.80$ volt] is:
  1. A 0.55 volt
  2. B - 1.05 volt
  3. C + 1.05 volt
  4. D - 0.55 volt
Show answer

Correct answer: C

9
The conductivity of four electrolytes $\mathrm { P } , \mathrm { Q } , \mathrm { R } , \mathrm { S }$ in ohm $\mathrm { cm } ^ { - 1 }$ are as follows $\mathrm { P } \left( 5 \times 10 ^ { - 5 } \right) , \mathrm { Q } \left( 1 \times 10 ^ { - 10 } \right)$, $\mathrm { R } \left( 7 \times 10 ^ { - 8 } \right) ; \mathrm { S } \left( 9.2 \times 10 ^ { - 3 } \right)$. The one which offers highest resistance to the passage of electric current is:
  1. A P
  2. B S
  3. C R
  4. D Q
Show answer

Correct answer: D

10
The activation energy of a reaction can be determined from the slope of which of the following graphs?
  1. A $\frac { \mathrm { T } } { \ln \mathrm { k } } \mathrm { Vs } \frac { 1 } { \mathrm {~T} }$
  2. B $\ln \mathrm { k } \mathrm { Vs } \mathrm { T }$
  3. C $\frac { \ln k } { T } \mathrm { Vs } T$
  4. D $\ln \mathrm { k } \mathrm { Vs } \frac { 1 } { \mathrm {~T} }$
Show answer

Correct answer: D

11
What is the charge on tin if 7.42 g of metallic tin is deposited by passage of 24125 C through a solution containing the ion? (Atomic weight of $\mathrm{Sn}=118 \mathrm{u}$ )
  1. A +1
  2. B +3
  3. C +2
  4. D +4
Show answer & explanation

Correct answer: D

$W=Z \times Q$
$7.42=\frac{118}{n \times 96500} \times 24125 \Rightarrow n=\frac{118}{7.42 \times 4}=4$
Therefore, option (4) is the correct answer.
12
Saturated solution of $\mathrm{KNO}_{3}$ is used to make salt bridge because:
  1. A velocity of $\mathrm{K}^{+}$is greater than that of $\mathrm{NO}_{3}^{-}$.
  2. B velocity of $\mathrm{NO}_{3}^{-}$is greater than that of $\mathrm{NO}_{3}^{-}$.
  3. C velocity of $\mathrm{K}^{+}$and $\mathrm{NO}_{3}^{-}$are nearly same.
  4. D $\mathrm{KNO}_{3}$ is highly soluble in water
Show answer & explanation

Correct answer: C

Velocity of $\mathrm{K}^{+}$and $\mathrm{NO}_{3}^{-}$are nearly same so that electrical neutrality is maintained.
Therefore, option (3) is the correct answer.
13
The standard emf of the cell is:
$\mathrm{Cr}\left|\mathrm{Cr}^{+3} \| \mathrm{Fe}^{+2}\right| \mathrm{Fe}$
$\mathrm{E}_{\mathrm{Cr}^{\circ} / \mathrm{Cr}^{3+}}^{\circ}=0.74 \mathrm{~V} ; \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V} ;$
  1. A 0.30 V
  2. B - 0.30 V
  3. C 1.18 V
  4. D - 1.18 V
Show answer

Correct answer: A

14
The equilibrium constant of the reaction:
$\mathrm { Cu } ( \mathrm { s } ) + 2 \mathrm { Ag } ^ { + } ( \mathrm { aq } ) \rightarrow \mathrm { Cu } ^ { 2 + } ( \mathrm { aq } ) + 2 \mathrm { Ag } ( \mathrm { s } ) ; \mathrm { E } ^ { \circ } = 0.46 \mathrm {V}$ at 298 K is;
  1. A $4.0 \times 10 ^ { 15 }$
  2. B $2.4 \times 10 ^ { 10 }$
  3. C $2.0 \times 10 ^ { 10 }$
  4. D $4.0 \times 10 ^ { 10 }$
Show answer

Correct answer: A

15
A hypothetical electrochemical cell is shown below $\mathrm { A } \left| \mathrm { A } ^ { + } ( \mathrm { xM } ) \| \mathrm { B } ^ { + } ( \mathrm { yM } ) \right| \mathrm { B }$, the emf measured is + 0.20 V . The cell reaction is:
  1. A $\mathrm { A } ^ { + } + \mathrm { B } \rightarrow \mathrm { A } + \mathrm { B } ^ { + }$.
  2. B $\mathrm { A } ^ { + } + \mathrm { e } ^ { - } \rightarrow \mathrm { A } ; \mathrm { B } ^ { + } + \mathrm { e } ^ { - } \rightarrow \mathrm { B }$.
  3. C the cell reaction cannot be predicted.
  4. D $\mathrm { A } + \mathrm { B } ^ { + } \rightarrow \mathrm { A } ^ { + } + \mathrm { B }$.
Show answer

Correct answer: D

16
From the following molar conductivities at infinite dilution,
$\begin{array} { l } \Lambda _ { \mathrm { m } } ^ { \circ } \text { for } \mathrm { Al } _ { 2 } \left( \mathrm { SO } _ { 4 } \right) _ { 3 } = 858 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 } \\ \Lambda _ { \mathrm { m } } ^ { \circ } \text { for } \mathrm { NH } _ { 4 } \mathrm { OH } = 238.3 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 } \\ \Lambda _ { \mathrm { m } } ^ { \circ } \text { for } \left( \mathrm { NH } _ { 4 } \right) _ { 2 } \mathrm { SO } _ { 4 } = 238.4 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 } \end{array}$
Calculate $\Lambda _ { \mathrm { m } } ^ { \circ }$ for $\mathrm { Al } ( \mathrm { OH } ) _ { 3 }$.
  1. A $715.2 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 }$
  2. B $1575.6 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 }$
  3. C $786.3 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 }$
  4. D $157.56 \mathrm {~S} \mathrm {~cm} ^ { 2 } \mathrm {~mol} ^ { - 1 }$
Show answer

Correct answer: C

17
A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by same current in 2 hours if current efficiency is $50 \%$ ?
  1. A 22 kg
  2. B 18 kg
  3. C 9 kg
  4. D 27 kg
Show answer & explanation

Correct answer: C

$\frac{\mathrm{W}_{\mathrm{Ca}}}{\mathrm{E}_{\mathrm{Ca}}}=\frac{\mathrm{W}_{\mathrm{Al}}}{\mathrm{E}_{\mathrm{Al}}} \Rightarrow \frac{40}{20}=\frac{\mathrm{W}_{\mathrm{Al}}}{9} \Rightarrow \mathrm{~W}_{\mathrm{Al}}=18 \mathrm{kg}$

If current efficiency is $100 \%$, then $\mathrm{W}_{A l}=18 \mathrm{~kg}$ Here, $\mathrm{W}_{\mathrm{Al}}=9 \mathrm{~kg}$ because current efficiency is $50 \%$. Therefore, option (3) is the correct answer.
18
$\mathrm{Cu}^{2+}\left(\mathrm{C}_{1}\right)+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}\left(\mathrm{C}_{2}\right)+\mathrm{Cu}(\mathrm{s})$
For the above cell reaction of an electrochemical cell, the free energy change $\Delta \mathrm{G}$ at a given temperature is a function of:
  1. A $\ln \mathrm{C}_{1} \times \mathrm{C}_{2}$
  2. B $\ln \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$
  3. C $\ln \mathrm{C}_{1}$
  4. D $\ln \mathrm{C}_{2}$
Show answer & explanation

Correct answer: B

$\Delta \mathrm{G}=-\mathrm{RT} \ln \mathrm{~K}$
$\Delta \mathrm{G}$ is a function of $\ln \mathrm{K}$ for a given temperature
$\mathrm{K}=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \Rightarrow \ln \mathrm{~K}=\ln \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$
19
The molar conductivity is maximum for the solution of concentration:
  1. A 0.001 M
  2. B 0.005 M
  3. C 0.002 M
  4. D 0.004 M
Show answer

Correct answer: A

20
Three faradays electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (Atomic weight $= 56 \mathrm { u }$) deposited at the cathode (in g) is:
  1. A 56
  2. B 84
  3. C 112
  4. D 168
Show answer

Correct answer: B

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