Electrochemistry — DPP 3

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
If the half cell reactions given as:
I. $\mathrm { Fe } ^ { 2 + } ( \mathrm { aq } ) + 2 \mathrm { e } ^ { - } \rightarrow \mathrm { Fe } ( \mathrm { s } ) , \mathrm { E } ^ { \circ } = - 0.44 \mathrm {~V}$
II. $2 \mathrm { H } ^ { + } ( \mathrm { aq } ) + \frac { 1 } { 2 } \mathrm { O } _ { 2 } ( \mathrm {~g} ) + 2 \mathrm { e } ^ { - } \rightarrow \mathrm { H } _ { 2 } \mathrm { O } _ { ( \mathrm { d } ) } , \mathrm { E } ^ { \circ } = + 1.23 \mathrm {~V}$
then $\mathrm { E } ^ { \circ }$ for the given reaction is; $\mathrm { Fe } ( \mathrm { s } ) + 2 \mathrm { H } ^ { + } + 1 / 2 \mathrm { O } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { Fe } ^ { 2 + } + \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { l } )$
  1. A + 1.67 V
  2. B - 1.67 V
  3. C + 0.79 V
  4. D - 0.79 V
Show answer

Correct answer: A

2
The relation between standard reduction potential of a cell and equilibrium constant is shown by:
  1. A $\mathrm { E } _ { \text {cell } } ^ { \circ } = \frac { \mathrm { n } } { 0.059 } \log \mathrm {~K} _ { \mathrm { c } }$
  2. B $\mathrm { E } _ { \text {cell } } ^ { \circ } = \frac { 0.059 } { \mathrm { n } } \log \mathrm { K } _ { \mathrm { c } }$
  3. C $\mathrm { E } _ { \text {cell } } ^ { \circ } = 0.059 \mathrm { n } \log \mathrm { K } _ { \mathrm { c } }$
  4. D $\mathrm { E } _ { \text {cell } } ^ { \circ } = \frac { \log \mathrm { K } _ { \mathrm { c } } } { \mathrm { n } }$
Show answer

Correct answer: B

3
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 $2.4 \times 10 ^ { 10 }$
  2. B $2.0 \times 10 ^ { 10 }$
  3. C $4.0 \times 10 ^ { 10 }$
  4. D $4.0 \times 10 ^ { 15 }$
Show answer

Correct answer: D

4
Mark the correct nernst equation for the given cell:
$\mathrm { Fe } ( \mathrm {s} ) \left| \mathrm { Fe } ^ { 2 + } ( \mathrm { aq } ) ( 0.001 \mathrm { M } ) \| \mathrm { H } ^ { + } ( \mathrm { aq } ) ( 1 \mathrm { M } ) \right| \mathrm { H } _ { 2 } ( \mathrm {g} ) ( 1 \mathrm { bar } ) \mid \mathrm { Pt } ( \mathrm {s} )$
  1. A $\mathrm { E } _ { \text {cell } } = \mathrm { E } _ { \text {cell } } ^ { \circ } - \frac { 0.0591 } { 2 } \log \frac { \left[ \mathrm { Fe } ^ { 2 + } \right] \left[ \mathrm { H } ^ { + } \right] ^ { 2 } } { [ \mathrm { Fe } ] \left[ \mathrm { H } _ { 2 } \right] }$
  2. B $\mathrm { E } _ { \text {cell } } = \mathrm { E } _ { \text {cell } } ^ { \circ } - \frac { 0.0591 } { 2 } \log \frac { \left[ \mathrm { H } ^ { + } \right] ^ { 2 } } { \left[ \mathrm { Fe } ^ { 2 + } \right] \left( \mathrm { p } _ { \mathrm { H } _ { 2 } } \right) }$
  3. C $\mathrm { E } _ { \text {cell } } = \mathrm { E } _ { \text {cell } } ^ { \circ } - \frac { 0.0591 } { 2 } \log \frac { \left[ \mathrm { Fe } ^ { 2 + } \right] \left( \mathrm { p } _ { \mathrm { H } _ { 2 } } \right) } { \left[ \mathrm { H } ^ { + } \right] ^ { 2 } }$
  4. D $\mathrm { E } _ { \text {cell } } = \mathrm { E } _ { \text {cell } } ^ { \circ } - \frac { 0.0591 } { 2 } \log \frac { \left( \mathrm { p } _ { \mathrm { H } _ { 2 } } \right) } { \left[ \mathrm { Fe } ^ { 2 + } \right] \left[ \mathrm { H } ^ { + } \right] ^ { 2 } }$
Show answer

Correct answer: C

5
The cell representation of the following reaction,
$\frac { 1 } { 2 } \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { AgCl } ( \mathrm { s } ) \longrightarrow \mathrm { H } _ { ( \mathrm { aq } ) } ^ { + } + \mathrm { Cl } _ { ( \mathrm { aq } ) } ^ { - } + \mathrm { Ag } ( \mathrm { s } )$, is:
  1. A $\mathrm { Ag } | \mathrm { AgCl } ( \mathrm { s } ) | \mathrm { KCl } ( \mathrm { aq } ) \| \mathrm { AgNO } _ { 3 } ( \mathrm { aq } ) \mid \mathrm { Ag } ( \mathrm { s } )$.
  2. B $\mathrm { Pt } \left| \mathrm { H } _ { 2 } ( \mathrm {~g} ) \right| \mathrm { HCl } ( \mathrm { ag } ) \| \mathrm { AgNO } _ { 3 } ( \mathrm { aq } ) \mid \mathrm { Ag } ( \mathrm { s } )$.
  3. C $\mathrm { Pt } \left| \mathrm { H } _ { 2 } ( \mathrm {~g} ) \right| \mathrm { HCl } ( \mathrm { aq } ) | \mathrm { AgCl } ( \mathrm { s } ) | \mathrm { Ag } ( \mathrm { s } )$.
  4. D $\mathrm { Pt } \left| \mathrm { H } _ { 2 } ( \mathrm {~g} ) \right| \mathrm { HCl } ( \mathrm { aq } ) | | \mathrm { KCl } ( \mathrm { aq } ) | \mathrm { AgCl } ( \mathrm { s } ) | \mathrm { Ag } ( \mathrm { s } )$.
Show answer

Correct answer: C

6
Calculate the emf of half cell:
$\left. \begin{array} { r | } \mathrm { Pt } , \mathrm { Cl } _ { 2 } ( \mathrm {~g} ) \\ ( 10 \mathrm { tm } ) \end{array} \right| _ { ( 0.1 ) \mathrm { M } } ^ { \mathrm { HCl } } ; \mathrm { E } _ { \mathrm { Cl } _ { 2 } / \mathrm { Cl } ^ { - } } ^ { \circ } = 1.36 \mathrm {~V}$
  1. A 1.45 V
  2. B 1.27 V
  3. C - 1.45 V
  4. D - 1.27 V
Show answer

Correct answer: A

7
The relationship between standard free energy and standard electrode potential is:
  1. A $\Delta \mathrm { G } ^ { \circ } = - \mathrm { nFE } ^ { \circ }$
  2. B $\Delta \mathrm { G } ^ { \circ } = \mathrm { nFE } ^ { \circ }$
  3. C $\Delta \mathrm { G } ^ { \circ } = \frac { \Delta \mathrm { H } } { \mathrm { nFE } ^ { \circ } }$
  4. D $\Delta \mathrm { G } ^ { \circ } = \frac { \mathrm { nFE } ^ { \circ } } { \mathrm { R } }$
Show answer

Correct answer: A

8
When Zn metal is added to $\mathrm { CuSO } _ { 4 }$ solution, Cu is precipitated. It is due to:
  1. A oxidation of $\mathrm { Cu } ^ { 2 + }$.
  2. B reduction of $\mathrm { Cu } ^ { 2 + }$.
  3. C hydrolysis of $\mathrm { CuSO } _ { 4 }$.
  4. D ionization of $\mathrm { CuSO } _ { 4 }$.
Show answer

Correct answer: B

9
Emf of hydrogen electrode in terms of pH is: [at 1atm pressure]
  1. A $\mathrm { E } _ { \mathrm { H } _ { 2 } } = \frac { \mathrm { RT } } { \mathrm { F } } \mathrm { pH }$
  2. B $\mathrm { E } _ { \mathrm { H } _ { 2 } } = \frac { \mathrm { RT } } { \mathrm { F } } \frac { 1 } { \mathrm { pH } }$
  3. C $\mathrm { E } _ { \mathrm { H } _ { 2 } } = \frac { 2.303 \mathrm {~T} } { \mathrm {~F} } \mathrm { pH }$
  4. D $\mathrm { E } _ { \mathrm { H } _ { 2 } } = - 0.0591 \mathrm { pH }$
Show answer

Correct answer: D

10
Agar-agar is used in salt bridge since it is a/an:
  1. A electrolyte.
  2. B non-electrolyte.
  3. C inert electrolyte.
  4. D a solid.
Show answer

Correct answer: B

11
Alkali metals have high oxidation potential, hence, they behave as good:
  1. A oxidising agent.
  2. B lewis base.
  3. C reducing agent.
  4. D electrolyte.
Show answer

Correct answer: C

12
Arrange the following in the order of their decreasing electrode potentials: $\mathrm { Mg } , \mathrm { K } , \mathrm { Ba }$ and Ca .
  1. A $\mathrm { K } > \mathrm { Ba } > \mathrm { Ca } > \mathrm { Mg }$
  2. B $\mathrm { Ba } > \mathrm { Ca } > \mathrm { K } > \mathrm { Mg }$
  3. C $\mathrm { Ca } > \mathrm { Mg } > \mathrm { K } > \mathrm { Ba }$
  4. D $\mathrm { Mg } > \mathrm { Ca } > \mathrm { Ba } > \mathrm { K }$
Show answer

Correct answer: D

13
If the solution of $\mathrm { CuSO } _ { 4 }$ in which copper rod is immersed is diluted to 10 times, then the electrode potential:
  1. A increases by 0.295 V
  2. B decreases by 0.0295 V
  3. C increases by 0.059 V
  4. D decreases by 0.059 V
Show answer

Correct answer: B

14
The purpose of the salt bridge in a galvanic cell is to:
  1. A prevent accumulation of charges around the electrodes.
  2. B facilitate continuity of the cell reaction.
  3. C to produce current at a constant strength.
  4. D All of these.
Show answer

Correct answer: D

15
$\Delta \mathrm { G } ^ { \circ }$ for the reaction, $\mathrm { Cu } ^ { 2 + } + \mathrm { Fe } \rightarrow \mathrm { Fe } ^ { 2 + } + \mathrm { Cu }$ is:
(Given: $\mathrm { E } _ { \mathrm { Cu } ^ { 2 + } / \mathrm { Ku } } ^ { \circ } = + 0.34 \mathrm {~V} , \mathrm { E } _ { \mathrm { Fe } ^ { 2 > } / \mathrm { Fe } } ^ { \circ } = - 0.44 \mathrm {~V}$ )
  1. A 11.44 KJ
  2. B 180.8 KJ
  3. C - 150.5 KJ
  4. D 28.5 KJ
Show answer

Correct answer: C

16
For a cell reaction involving a two electron exchange, the standard emf of the cell is found to be 0.295 V at $25 ^ { \circ } \mathrm { C }$. The equilibrium constant of the reaction at $25 ^ { \circ } \mathrm { C }$ will be:
  1. A $1 \times 10 ^ { - 10 }$
  2. B $29.5 \times 10 ^ { - 2 }$
  3. C 10
  4. D $1 \times 10 ^ { 10 }$
Show answer

Correct answer: D

17
If a salt bridge is not used between two half cells, voltage:
  1. A drops to zero
  2. B does not change.
  3. C increases gradually
  4. D increases rapidly
Show answer

Correct answer: A

18
The electrode potentials for:
$\mathrm { Cu } ^ { 2 + } ( \mathrm { aq } ) + \mathrm { e } ^ { - } \rightarrow \mathrm { Cu } ^ { + } ( \mathrm { aq } )$
and $\mathrm { Cu } ^ { + } ( \mathrm { aq } ) + \mathrm { e } ^ { - } \rightarrow \mathrm { Cu } ( \mathrm { aq } )$
are 0.15 V and + 0.50 V respectively. The value of $\mathrm { E } _ { \mathrm { Cu } ^ { 2 + } / \mathrm { Cu } } ^ { \circ }$ will be:
  1. A 0.150 V
  2. B 0.500 V
  3. C 0.325 V
  4. D 0.650 V
Show answer

Correct answer: C

19
The standard emf of a galvanic cell with electrode potential of $\mathrm { Zn } ^ { 2 + } / \mathrm { Zn } = - 0.76 \mathrm {~V}$ and that of $\mathrm { Cu } ^ { 2 + } / \mathrm { Cu } = 0.34 \mathrm {~V}$, is:
  1. A 0.76 V
  2. B 1.1 V
  3. C - 1.1 V
  4. D 1.6 V
Show answer

Correct answer: B

20
The electrode potential measures the:
  1. A tendency of the electrode to gain or lose electrons.
  2. B electron affinity of elements.
  3. C difference in the ionization potential of electrode and metal ion.
  4. D heat of combustion.
Show answer

Correct answer: A

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