CBSE Class 12 Chemistry 2019 Delhi Set 1 Solved Paper

Question : 25 of 27

Marks: +1, -0

SECTION - D

{\E}°_{\;\text "cell "\;}

for the given redox reaction is

2.71 {\V}

{\Mg}+ {\Cu}^{2+}{ }_{(0.01 {\M})} → {\Mg}^{2+}{ }_{(0.001 {\M})}+ {\Cu}

Calculate

E_{\;\text "cell "\;}

for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) Less than

2.71 {\V}

(ii) Greater than

2.71 {\V}

OR
(a) A steady current of 2 amperes was passed through two electrolytic cells

{\X}

and

{\Y}

connected in series containing electrolytes

{\FeSO}_4

and

{\ZnSO}_4

until 2.8

{\g}

of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass

\;\;\text " (Molar mass: Fe "\;=56 {\g} {\mol}^{-1}, {\Zn}

=65.3 {\g} {\mol}^{-1}, 1 {\F}

\;=96500 {\C} {\mol}^{-1})

(b) In the plot of molar conductivity

\Λ_{ {\m}}

vs. square root of concentration

( {\C}^{1 \/ 2})

, following curve obtained for two electrolytes A and B:

Answer the following :
(i) Predict the nature of electrolytes A and B:
(ii) What happens on extrapolation of

\Λ_m

to concentration approaching zero for electrolytes

A

and

B

Solution:

\; {\E}_{\;\text "cell "\;}= {\E}°_{\;\text "cell "\;} -\;{0.059}/{ {\n}} \log \;{ [ {\Mg}^{2+}] }/{ [ {\Cu}^{2+}] }

\; {\E}_{\;\text "cell "\;}=2.71-\;{0.059}/{2} \log \;{[0.001]}/{[0.01]}

\; {\E}_{\;\text "cell "\;}=2.71-(-0.0295)=2.74 {\V}

(i) When external opposite applied voltage is less than 2.71, it is less than

E°_{\;\text "cell "\;}

, therefore, the electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).

(ii) When external opposite applied potential is greater than 2.71, it is greater than

E°_{\;\text "cell "\;}

, therefore, the reaction will be reversed, and the current will flow from anode to cathode.
OR
(a) Charge required to deposite

2.8 {\g} {\Fe}

{\mol\ \Fe}=\;{\;\text " mass "\;}/{\;\text " molar mass "\;}

=\;{2.8 {\g}}/{56 {\g} ⋅ {\mol}^{-1}}=0.05 {\mol}

2 {\F}

charge is required to discharge

1 {\mol}

{\Fe}^{2+}

ions as

{\Fe}

, therefore deposition of

0.05 {\mol} {\Fe}

will need

0.05 × 2=0.1 {\F}=0.1 {\F}

× \;{96500 {\C}}/{ {\F}}=9650 {\C}

The quantity of charge is related to current as

{\Q}= {\It}

Therefore, the time needed to deposit

2.8 {\g}

Fe is:

{\t}=\;{ {\Q}}/{ {\I}}=\;{9650 {\C}}/{2 {\A}}=4825 {\ s}

So, the current flowed through the cells for 4825 seconds.

The amount of

{\Zn}

deposited in cell

{\Y}

can be calculated using Faraday's second law:

\;\;{\;\text " mass of "\; {\Zn}}/{\;\text " mass of "\; {\Fe}}=\;{\;\text " Eq.wt of "\; {\Zn}}/{\;\text " Eq wt of "\; {\Fe}}

\; \;\; =\;{\;\text " molar mass of "\; {\Zn} \/ \;\text " chargeon zincion "\;}/{\;\text " molarmass of "\; {\Fe} \/ \;\text " chargeon iron ion "\;}

\;\;\text " mass of "\; {\Zn}=2.8 {\g} × \;{65.3 {\g} \/ 2}/{56 {\g} \/ 2}

=3.265 {\g} ≈ 3.3 {\g}

Therefore, the mass of

{\Zn}

deposited in cell

{\Y}

in the same time is

3.3 {\g}

.

(b) (i) Molar conductivity of strong electrolytes increases linearly as the square root of the concentration decreases; therefore, electrolyte A is a strong electrolyte. Molar conductivity of weak electrolytes increases non-linearly as square root of concentration decreases; therefore, electrolyte B is a weak electrolyte.
(ii) As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the molar conductivity axis, giving the limiting value of molar conductivity

E_m^0

. The plot of molar conductivity of weak electrolyte tends to infinity as its concentration approaches zero; it does not intersect the molar conductivity axis.

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