Electrochemistry Questions — NEET Chemistry

Q1

A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is : (Given : Molar mass of

\mathrm{Cu}=63 \mathrm{~g} \mathrm{~mol}^{-1}

;

\left.1 \mathrm{~F}=96487 \mathrm{C} \mathrm{~mol}^{-1}\right)

A 1.7018 g

B 0.2938 g

C 2.4036 g

D 0.5876 g

Correct Answer

Option B

Solution

When copper sulphate ( $\mathrm{CuSO}_4$ ) is dissolved in water, it splits into ions as follows:

\mathrm{CuSO}_4 \longrightarrow \mathrm{Cu}^{2+} + \mathrm{SO}_4^{2-}

During electrolysis, copper ions move towards the cathode (negative electrode) where they gain electrons and get deposited as solid copper metal:

\mathrm{Cu}^{2+} + 2e^- \longrightarrow \mathrm{Cu(s)}

According to Faraday’s laws of electrolysis, the mass of a substance deposited is directly proportional to the quantity of charge passed.

The mathematical form is:

W = \frac{E \times I \times t}{96500}

where, $W$ = mass of substance deposited, $E$ = electrochemical equivalent (equal to molar mass / number of electrons involved), $I$ = current in amperes, $t$ = time in seconds, and $96500\, \mathrm{C/mol}$ = 1 Faraday.

For copper, one atom needs 2 electrons, so $E = \dfrac{63}{2}$ .

Now substituting the given values: $I = 1.5\,\mathrm{A}$ , $t = 10 \times 60 = 600\,\mathrm{s}$

W = \frac{63 \times 1.5 \times 600}{2 \times 96500}

Simplifying,

W = \frac{56,700}{193000} = 0.2938\,\mathrm{g}

Hence, the mass of copper deposited on the cathode is $0.2938\,\mathrm{g} \approx 0.29\,\mathrm{g}$ .

Q2

From the following select the one which is not an example of corrosion.

A Rusting of iron object

B Production of hydrogen by electrolysis of water

C Tarnishing of silver

D Development of green coating on copper and bronze ornaments

Correct Answer

Option B

Solution

Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal.

The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion.

$\bullet$ Production of H

_2

by electrolysis of water is an example of electrolytic cell.

Q3

The standard cell potential of the following cell

\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right| \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Fe}

is

0.32 \mathrm{~V}

. Calculate the standard Gibbs energy change for the reaction:

\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})

(Given :

1 \mathrm{~F}=96487 \mathrm{C}

)

A

-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}

B

+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}

C

-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}

D

+61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct Answer

Option A

Solution

To find the standard Gibbs energy change for the reaction, we can use the relationship between the standard Gibbs free energy change (

\Delta G^\circ

) and the standard cell potential (

E^\circ

) given by the following equation:

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

Where:

n

is the number of moles of electrons transferred in the reaction.

F

is the Faraday constant (

1 \mathrm{~F} = 96487 \mathrm{~C \cdot mol^{-1}}

).

E^\circ

is the standard cell potential. From the given reaction:

\mathrm{Zn}(\mathrm{s}) + \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{Fe}(\mathrm{s})

It can be observed that

n = 2

because two electrons are transferred from zinc to iron. Given data:

E^\circ = 0.32 \mathrm{~V}

1 \mathrm{~F} = 96487 \mathrm{C/mol}

Now, substitute the values into the equation:

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

\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}

Calculate the value:

\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}

\Delta G^\circ = -2 \times 30875.84 \mathrm{~J/mol}

\Delta G^\circ = -61751.68 \mathrm{~J/mol}

Converting the units to kJ/mol:

\Delta G^\circ = -61.75168 \mathrm{~kJ/mol}

The closest answer, when rounded to two decimal places, is:

\Delta G^\circ = -61.75 \mathrm{~kJ/mol}

Therefore, the correct option is: Option A:

-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}

Q4

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of

\mathrm{Cu}: 63 \mathrm{~g} \mathrm{~mol}^{-1}, 1 \mathrm{~F}=96487 \mathrm{C}

)

A 3.15 g

B 0.315 g

C 31.5 g

D 0.0315 g

Correct Answer

Option B

Solution

The mass of copper deposited when passing an electric current through a copper sulphate solution can be calculated using Faraday's laws of electrolysis.

The first law states that the amount of a substance deposited or liberated at an electrode during electrolysis is proportional to the amount of electricity (charge) passed through the electrolyte.

To find the mass of copper deposited, we use the formula:

m = \frac{M \times I \times t}{n \times F}

Where:

m

is the mass of the substance deposited (in grams),

M

is the molar mass of the substance (in g/mol),

I

is the current (in amperes, A),

t

is the time electricity is passed through the solution (in seconds),

n

is the number of moles of electrons required to deposit or dissolve 1 mole of the substance (valence number, which is 2 for copper in copper sulphate solution as copper ions are

\mathrm{Cu^{2+}}

),

F

is the Faraday constant, approximately

96487 \mathrm{C/mol}

of electrons. Given:

M = 63 \mathrm{~g/mol}

I = 9.6487 \mathrm{~A}

t = 100 \mathrm{~s}

n = 2

F = 96487 \mathrm{~C/mol}

Substitute these values into the formula:

m = \frac{63 \mathrm{~g/mol} \times 9.6487 \mathrm{~A} \times 100 \mathrm{~s}}{2 \times 96487 \mathrm{~C/mol}}

Calculate the value of

m

:

m = \frac{63 \times 9.6487 \times 100}{2 \times 96487}

m = \frac{60828.21}{192974}

m \approx 0.315 \mathrm{~g}

Hence, the mass of copper deposited is approximately 0.315 g, making Option B the correct answer.

Q5

The

\mathrm{E}^{\Theta}

values for

\begin{aligned} & \mathrm{Al}^{+} / \mathrm{Al}=+0.55 \mathrm{~V} \text{ and } \mathrm{Tl}^{+} / \mathrm{Tl}=-0.34 \mathrm{~V} \\ & \mathrm{Al}^{3+} / \mathrm{Al}=-1.66 \mathrm{~V} \text{ and } \mathrm{T}^{3+} / \mathrm{Tl}=+1.26 \mathrm{~V} \end{aligned}

Identify the incorrect statement

A

\mathrm{Al}

is more electropositive than

\mathrm{Tl}

B

\mathrm{Tl}^{3+}

is a good reducing agent than

\mathrm{Tl}^{1+}

C

\mathrm{Al}^{+}

is unstable in solution

D

\mathrm{Tl}

can be easily oxidised to

\mathrm{Tl}^{+}

than

\mathrm{Tl}^{3+}

Correct Answer

Option B

Solution

\mathrm{Tl}^3

act as an oxidising agent not reducing agent.

Q6

The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is

\begin{aligned} & \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \text{ (aq) }+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}^{\circ}=+1.33 \mathrm{~V} \end{aligned}

A

+1.77 \mathrm{~V}

B

+2.65 \mathrm{~V}

C

+0.01 \mathrm{~V}

D

+0.89 \mathrm{~V}

Correct Answer

Option A

Solution

\begin{aligned} \mathrm{E}_{\text{cell }}^{\circ} & =\mathrm{E}_{\mathrm{C}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{A}}^{\mathrm{o}} \\ & =(1.33)-(-0.44) \\ & =+1.77 \mathrm{~V} \end{aligned}

Q7

The conductivity of centimolar solution of

\mathrm{KCl}

at

25^{\circ} \mathrm{C}

is

0.0210 ~\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}

and the resistance of the cell containing the solution at

25^{\circ} \mathrm{C}

is

60 ~\mathrm{ohm}

. The value of cell constant is -

A

3.28 \mathrm{~cm}^{-1}

B

1.26 \mathrm{~cm}^{-1}

C

3.34 \mathrm{~cm}^{-1}

D

1.34 \mathrm{~cm}^{-1}

Correct Answer

Option B

Solution

Centimolar solution

=\frac{1}{100} \mathrm{M}=0.01 ~\mathrm{M}

Conductivity

(k)=0.0210 ~\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}

Resistance

(\mathrm{R})=60 ~\mathrm{ohm}

k=\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{A}}\right)

\Rightarrow 0.0210=\frac{1}{60}\left(\frac{\ell}{\mathrm{A}}\right) \Rightarrow \frac{\ell}{\mathrm{A}}=1.26 \mathrm{~cm}^{-1}

Q8

Standard electrode potential for the cell with cell reaction Zn(s) + Cu 2+ (aq)

\to

Zn 2+ (aq) + Cu(s) is 1.1 V. Calculate the standard Gibbs energy change for the cell reaction. (Given F = 96487 C mol

-

1 )

A

-

200.27 J mol

-

1

B

-

200.27 kJ mol

-

1

C

-

212.27 kJ mol

-

1

D

-

212.27 J mol

-

1

Correct Answer

Option C

Solution

Zn(s) + Cu 2+ (aq) $\to$ Zn 2+ (aq) + Cu(s) E

_{cell}^0

= 1.1 V

\Delta

G

^\circ

= $-$ nFE

_{cell}^0

$\therefore$ n = 2

\Delta

G

^\circ

= $-$ 2 $\times$ 96487 $\times$ 1.1

\Delta

G

^\circ

= $-$ 212271.4 J mol $-$ 1

\Delta

G

^\circ

= $-$ 212.27 kJ mol $-$ 1

Q9

At 298 K, the standard electrode potentials of Cu 2+ / Cu, Zn 2+ / Zn, Fe 2+ / Fe and Ag + / Ag are 0.34 V,

-

0.76 V,

-

0.44 V V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

A CuSO 4 (aq) + Zn(s)

\to

ZnSO 4 (aq) + Cu(s)

B CuSO 4 (aq) + Fe(s)

\to

FeSO 4 (aq) + Cu(s)

C FeSO 4 (aq) + Zn(s)

\to

ZnSO 4 (aq) + Fe(s)

D 2CuSO 4 (aq) + 2Ag(s)

\to

2Cu(s) + Ag 2 SO 4 (aq)

Correct Answer

Option D

Solution

For a reaction to be spontaneous, E

_{cell}^o

must be positive. $\bullet$ For, FeSO 4 (aq) + Zn(s) $\to$ ZnSO 4 (aq) + Fe(s) E

_{cell}^o

= E

_{cathode}^o

$-$ E

_{anode}^o

= $-$ 0.44 V $-$ ( $-$ 0.76 V) = 0.32 V $\bullet$ For, 2CuSO 4 (aq) + 2Ag(s) $\to$ 2Cu(s) + Ag 2 SO 4 (aq) E

_{cell}^o

= 0.34 V $-$ 0.80 V = $-$ 0.46 V $\bullet$ For, CuSO 4 (aq) + Zn(s) $\to$ ZnSO 4 (aq) + Cu(s) E

_{cell}^o

= 0.34 V $-$ ( $-$ 0.76 V) = 1.1 V $\bullet$ For, CuSO 4 (aq) + Fe(s) $\to$ FeSO 4 (aq) + Cu(s) E

_{cell}^o

= 0.80 V $-$ ( $-$ 0.44 V) = 1.24 V

Q10

Find the emf of the cell in which the following reaction takes place at 298 K Ni(s) + 2Ag + (0.001 M)

\to

Ni 2+ (0.001 M) + 2Ag(s) (Given that E

_{cell}^o

= 10.5 V,

{{2.303\,RT} \over F} = 0.059

at 298 K)

A 10.4115 V

B 10.385 V

C 0.9615 V

D 10.05 V

Correct Answer

Option A

Solution

To determine the electromotive force (EMF) of the given cell reaction at 298 K: Reaction: Ni(s) + 2Ag + (0.001 M) → Ni 2+ (0.001 M) + 2Ag(s) Given data: Standard cell potential, E° cell = 10.5 V Value at 298 K,

\frac{2.303\,RT}{F}

= 0.059 Using the Nernst equation to find the cell potential, E cell : $E_{cell} = E_{cell}^o - \dfrac{0.059}{n} \log \dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$ Where: n = number of moles of electrons transferred = 2 $[\text{Ni}^{2+}] = 0.001\, \text{M}$ $[\text{Ag}^+] = 0.001\, \text{M}$ Substituting the values: $E_{cell} = 10.5 - \dfrac{0.059}{2} \log \dfrac{(10^{-3})}{(10^{-3})^2}$ $= 10.5 - \dfrac{0.0295}{2} \log (10^3)$ $= 10.5 - 0.0295 \times 3$ $= 10.5 - 0.0885$ $= 10.4115\, \text{V}$ Therefore, the EMF of the cell is 10.4115 V.