Important Questions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry

Question 1. How would you determine the standard electrode potential of the system Mg²⁺/Mg?
Answer:

The standard electrode potential of Mg²⁺ | Mg can be measured concerning the standard hydrogen electrode, represented by Pt(s). H₂(g) (1 atm) | H⁺(aq) (1 M). A cell, consisting of Mg I MgSO₄. (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

⇒ \(\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{H}(\mathrm{g}, 1 \text { bar), } \mathrm{Pt}(\mathrm{s})\)

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

⇒ \(E^{(-)}=E_R^{(-)}-E_L^{(-)}\)

Here, the E^–_R for the standard hydrogen electrode is zero.

Therefore, E^– = 0 – E^–_L = E^–_L

Question 2. Can you store copper sulphate solution in a zinc pot?
Answer:

No. Zinc is more reactive than copper. Zinc reacts with copper sulphate and displaces copper. Zn + CuSO₄+ ZnSO₄ + Cu

Question 3. Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Answer:

⇒ \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\)

⇒ \(E_{\mathrm{oxi}}^0=-0.77 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{red}}^0=+0.77 \mathrm{~V}\)

Substances that are stronger oxidizing agents and have greater reduction potential than +0.77 V will oxidize

Fe²⁺ for example., Br₂, Cl₂ and F₂

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Answer:

2H⁺ + 2e^– → H₂

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\mathrm{E}_{\mathrm{cell}}^o=0, \mathrm{P}_{\mathrm{H}_2}=1 \mathrm{~atm},\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-10}\)

⇒ \(E_{\text {cell }}=-\frac{0.059}{2} \log \frac{1}{\left(10^{-10}\right)^2}=-0.59 \mathrm{~V}\)

Question 5. Calculate the emf of the cell in which the following reaction takes place:

Nit(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s)

Given that \(E_{\text {cell }}^{\circ}=1.05 \mathrm{~V}\)

Answer:

Applying the Nernst equation we have:

⇒ \(E_{\text {(cell) }}=E_{\text {cell }}^{(-)}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)

⇒ \(1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^2}\)

⇒ \(1.05-0.02955 \log \frac{0.16}{0.000004}\)

= 1.05 -0.02955 log 4 x 104

= 1.05 – 0.02955 (log 10000 + log 4)

= 1.05 – 0.02955 (4 + 0.6021)

= 1.05 -0.02955 (4.6021)

= 1.05-0.14 = 0.91 V

Question 6. The cell in which the following reaction occurs:

⇒ \(\left.2 \mathrm{Fe}^{3+} \text { (aq. }\right)+2 \mathrm{I}^{-} \text {(aq.) } \rightarrow 2 \mathrm{Fe}^{2+} \text { (aq.) }+\mathrm{I}_2 \text { (s) has } \mathrm{E}_{\text {cell }}^o=0.236 \mathrm{~V} \text { at } 298 \mathrm{~K}\)

Calculate the standard Gibbs energy and equilibrium constant of the cell reaction.

Answer:

⇒ \(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}=-2 \times 96500 \times 0.236 \text { joule } \mathrm{mol}^{-1}=-45.55 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Δ_rG^o = -2.303 RT log K_c

(R = 8.314 J K^-1 mol^-1)

-45.55 = -2.303 x 8.314 x 298 log K_c

or log K_c = 7.981

K_c = antilog (7.98 1)

or K_c = 9.6 x 10⁷

Question 7. Why does the conductivity the a solution decrease with dilution?
Answer:

The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) per unit volume decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

Question 8. Suggest a way to determine the Δ^o_m, value of water.
Answer:

Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows :

Δ⁰_m(H2O) = Δ⁰_H=Δ_OH^–

⇒\(\left(\Delta_{\mathrm{H}^{+}}^0+\Delta_{\mathrm{Cl}^{-}}^0\right)+\left(\Delta_{\mathrm{Na}}^0+\Delta_{\mathrm{OH}^{-}}^0\right)-\left(\Delta_{\mathrm{Na}^{-}}^0+\Delta_{\mathrm{Cl}^{-}}^0\right)\)

⇒ \(\Delta_{\mathrm{m}\left(\mathrm{H}_2 \mathrm{O}\right)}^0=\Delta_{\mathrm{m}(\mathrm{HCl})}^0+\Delta_{\mathrm{m}(\mathrm{NaOH})}^0-\Delta_{\mathrm{m}(\mathrm{NaCl})}^0\)

Hence, by knowing the values of HCl, NaOH. and NaCl, the value of water can be determined.

Question 9. The molar conductivity of 0.025 mol/L methanoic acid is 46.1 S cm²/mol. Calculate the degree of dissociation and dissociation constant. Given: λ⁰_H = 349.6 and λ⁰_HCCO^– = 54.6

λ⁰_HCCO =λ⁰_H + λ⁰_HCCO^– = 349.6 + 54.6 = 404.2

⇒ \(\alpha=\frac{\lambda_{\mathrm{m}}^{\mathrm{c}}}{\lambda_{\mathrm{m}}^{\mathrm{o}}}=\frac{46.1}{404.2}=0.114\)

⇒ \(K_a=\frac{C \alpha^2}{1-\alpha}=\frac{0.025 \times 0.114 \times 0.114}{1-0.114}=3.67 \times 10^{-4}\)

Question 10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Answer:

I = 0.5 A. t = 2 hours = 2 × 60 × 60s = 7200 s Q = It = 0.5A × 7200 s = 3600C

∵ 96500 coulomb are equivalent of 6.023 x 10²³ number of electrons

∴ 3600 coulomb are equivalent to \(\frac{6.023 \times 10^{23} \times 3600}{96500}\)

= 2.246 x 10²² electrons

Question 11. Suggest a list of metals that are extracted electrolytically.
Answer: Alkali metals such as Na, K, etc., alkaline earth metals such as Mg, Ca, etc., and aluminum.

Question 12. Consider the reaction, Cr₂O^2-₇+ 14H⁺+ 6e^– → 2Cr³⁺ +7H₂O What is the quantity of electricity in coulombs needed to reduce one mole of Cr₂O^2-₇?
Answer:

To reduce 1 mole of Cr₂O^2-₇. the required quantity of electricity will be:

6F = 6 x 96487 coulomb = 578922 C

Question 13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:

lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO,v) as the cathode, and a 38% solution of sulphuric acid (H₂SO₄) as an electrolyte.

When the battery is in use, the following cell reactions take place:

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging, PbSO₄ (s) present at the anode and cathode is converted into Pb(s) and PbO₂(s) respectively.

Question 14. Suggest two materials other than hydrogen (that can be used as fuels in fuel cells.
Answer: Methane and methanol.

Question 1. Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg, and Zn

Answer:

The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Ah Zn, Fe, Cu

Question 2. Given the standard electrode potentials,

1. K⁺/K = -2.93V, Ag⁺/Ag = 0.80 V
2. Hg²⁺/Hg = 0.79V
3. Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = – 0.74V A

Arrange these metals in their increasing order of reducing power.
Answer:

The lower the reduction potential, the higher the reducing power.

The given standard electrode potentials increase in the order of

K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Question 3. Depict the galvanic cell in which the reaction \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) takes place. Further show:

1. Which of the electrodes is negatively charged?
2. The carriers of the current in the cell?
3. Individual reaction at each electrode.

Answer:

The galvanic cell in which the given reaction takes place is depicted as:

⇒ \(\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}^{2+}{ }_{\text {(aq) }} \| \mathrm{Ag}_{(\mathrm{aq})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\)

Zn electrode (anode) is negatively charged.

Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

The reaction taking place at the anode is given by,

⇒ \(\mathrm{Zn}_{(\mathrm{s})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)

The reaction taking place at the cathode is given by,

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(\mathrm{s})}\)

Question 4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place :

⇒ \(2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Cd}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Cd}(\mathrm{s})\)

⇒ \(\mathrm{E}_{\mathrm{Cr}^3 / \mathrm{Cr}}^{\ominus}=-0.74 \mathrm{~V} ; \quad E_{\mathrm{Cd}^2 / \mathrm{Cd}}^{\ominus}=-0.40 \mathrm{~V}\)

⇒ \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\)

⇒ \(\mathrm{E}_{\mathrm{Fe}^+3 / \mathrm{Fe}^+2}^{\ominus}=-0.77 \mathrm{~V} ; \quad E_{\mathrm{Ag}^+ / \mathrm{Ag}}^{\ominus}=-0.80 \mathrm{~V}\)

Calculate the Δ_rG^– and equilibrium constant of the reactions.

Answer:

The galvanic cell of the given reaction is depicted as:

⇒ \(\mathrm{Cr}_{(\mathrm{s})}\left|\mathrm{Cr}^{3+}{ }_{(\mathrm{aq})} \| \mathrm{Cd}^{2+}{ }_{(\mathrm{at})}\right| \mathrm{Cd}_{(\mathrm{s})}\)

Now, the standard cell potential is

⇒ \(\mathrm{E}_{\text {cell }}^{\ominus}=\mathrm{E}_{\mathrm{R}}^{\ominus}-\mathrm{E}_{\mathrm{L}}^{\ominus}\)

= -0.40- (-0.74) = +0.34 V

⇒ \(\Delta_r G^{\ominus}=-n F E_{\text {cell }}^{\ominus}\)

In the given equation,

n = 6

F = 96487 C mol^-1

⇒ \(\mathrm{E}_{\text {cell }}^{\ominus}=+0.34 \mathrm{~V}\)

Then, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-6 \times 96487 \mathrm{C} \mathrm{mol}^{-1} \times 0.34 \mathrm{~V}\)

= -196833.48 CV mol^-1

= -196833.48 J mol^-1

= -196.83 kJ mol^-1

Again, \(\Delta_r \mathrm{G}^{\ominus}=-\mathrm{RT} \ln \mathrm{K}\)

⇒ \(\Delta_r G^{\ominus}=-2.303 R T \log K\)

⇒ \(\log K=-\frac{\Delta_{\mathrm{r}} \mathrm{G}}{2.303 \mathrm{RT}}=\frac{-196.83 \times 10^3}{2.303 \times 8.314 \times 298}=34.496\)

K = antilog (34.496) = 3.13 x 10 34

⇒ \(\mathrm{E}_{\mathrm{Fe}^+3 / \mathrm{Fe}^+2}^{\ominus}=-0.77 \mathrm{~V} ; \quad E_{\mathrm{Ag}^+ / \mathrm{Ag}}^{\ominus}=-0.80 \mathrm{~V}\)

The galvanic cell of the given reaction is depicted as:

⇒ \(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\left|\mathrm{Fe}_{(\mathrm{aq})}^{3+}\right| \| \mathrm{Ag}_{(\mathrm{aq})}^{+} \mid \mathrm{Ag}_{(\mathrm{s})}\)

Now, the standard ceil potential is

⇒ \(\mathrm{E}_{\mathrm{cell}}^{\ominus}=\mathrm{E}_{\mathrm{R}}^{\ominus}-\mathrm{E}_{\mathrm{L}}^{\ominus}\)

= 0.80- 0.77- 0.03 V

Here, n = 1,

Then, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-n F E_{\mathrm{cell}}^{\ominus}\)

= -1 x 96487 C mor’ x 0.03 V.

= -2894,61 J mol-1

= -2.89 kJ mol-1

Again, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=2.303 \mathrm{RT}{Iog}\mathrm{k}\)

⇒ \(\log K=\frac{\Delta_r G}{2.303 R T}=\frac{-2894.61}{2.303 \times 8.314 \times 298}=0.5073\)

∴ K = antilog (0.5073) = 3.2 (approximately)

5. Write the Nernst equation and emf of the following cells at 298K:

⇒ \({Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\)

⇒ \(\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_2(\mathrm{~g})(\text { lbar }) \mid \mathrm{Pt}(\mathrm{s})\)

⇒ \({Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.020 \mathrm{M})\right| \mathrm{H}_2(\mathrm{~g}) \text { (Ibar) } \mid \mathrm{Pt}(\mathrm{s})\)

⇒ \({Pt}(\mathrm{s})\left|\mathrm{Br}_2(1)\right| \mathrm{Br}^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M})\left|\mathrm{H}_2(\mathrm{~g})(1 \mathrm{bar})\right| \mathrm{Pt}(\mathrm{s})\)

Answer:

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\ominus}-\frac{0.0591}{\mathrm{n}} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)

⇒ \(\{0.34-(-2.36)\}-\frac{0.0591}{2} \log \frac{.001}{.0001}\)

⇒ \(2.7-\frac{0.0591}{2} \log 10\)

⇒ 2.7 -0.02955

⇒ 2.67 (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\ominus}-\frac{0.0591}{\mathrm{n}} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\{0-(-0.44)\}-\frac{0.0591}{2} \log \frac{0.001}{1^2}\)

⇒ 0.44- 0.02955(-3)

⇒ 0.52865

⇒ 0.53 V (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\{0-(-0.14)\}-\frac{0.0591}{2} \log \frac{0.050}{(0.020)^2}\)

⇒ 0.14-0.0295 x log 125

= 0.14-0.062

= 0.078 V

= 0.08 V (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\odot}-\frac{0.0591}{\mathrm{n}} \log \frac{1}{\left[\mathrm{Br}^{-}\right]^2\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(=(0-1.09)-\frac{0.0591}{2} \log \frac{1}{(0.010)^2(0.030)^2}\)

⇒ \(1.09-0.02955 \times \log \frac{1}{0.00000009}\)

⇒ \(1.09-0.02955 \times \log \frac{1}{9 \times 10^{-8}}\)

⇒ 1.09-0.02955 x log(1.11 x 107)

⇒ -1.09-0.02955 (0.0453 + 7)

⇒ -1.09-0.208

⇒ -1.298 V

Question 6. In the button cells widely used in wanted and other devices, the following reaction takes place:

⇒ \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2(\mathrm{O})(l) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})\)

Determine Δ_rG^o and E^– for the reaction.

⇒ \(\mathrm{Zn}_{(\mathrm{s})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)

⇒ \(\mathrm{E}^{\ominus}=0.76 \mathrm{~V}\)

⇒ \(\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{l}_{(0)}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}_{(\mathrm{g})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} ; \mathrm{E}^{\ominus}=0 .{344 \mathrm{~V}}\)

⇒ \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{f})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} ; \mathrm{E}^{\ominus}=1.104 \mathrm{~V}\)

∴ E^– = 1.104 V

We know that,

⇒ \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-\mathrm{nF} \mathrm{E}^{\ominus}\)

= -2 × 96487 × 1.04

= -213043.296 J

= -213.04 kJ

Question 7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

⇒ \(\mathrm{G}=\kappa \frac{\mathrm{A}}{\mathrm{l}}=\kappa \cdot 1=\kappa \quad(\text { Since } \mathrm{A}=1,=1)\)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume (that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length is 1.

⇒ \(\Delta_{\mathrm{m}}=\kappa \frac{\mathrm{A}}{1}\)

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

⇒ \(\Lambda_m=\kappa V\)

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of Δ_m with √c for strong and weak electrolytes is shown in the following plot:

Question 8. The conductivity of the 0.20 M solution of KCI at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.
Answer:

Given, K = 0.0248 S cm^-1: c = 0.20 M

∴ Molar conductivity,

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa \times 1000}{c}=\frac{0.0248 \times 1000}{0.2}=124 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)

Question 9. The resistance of a conductivity cell containing 0.00 1 M KCI solution at 298 K is 1500Ω. What is the cell constant if the conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10^-3 S cm^-1?
Answer:

Given, Conductivity, K = 0.146 x 10^-3 S cm^-1

Resistance, R = 1500Ω

∴ Cell constant =K × R

= 0. 1 46 × 10^-3 × 1500 = 0.219 cm^-1

Question 10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and are results are given below:

Calculate Δ_m for all concentrations and draw a plot between Δ_m and C1/2. Find the value of Δ^o_m

Answer:

Given,

K = 1.237 × 10^-2 Sm^-1, c = 0.001 M

Then, K = 1.237 × 10^-1 S cm^-1 c^1/2= 0.03 1 6 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa}{\mathrm{c}}=\frac{1.237 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.001 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=123.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

K = 1185 × 10^-2 s m^-1, c =0.010 M

Then, ic = 11.85 x 10^-4 S cm^-1, c^1/2 = 0.1 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{c}}=\frac{11.85 \times 10^{-4} \mathrm{Scm}^{-1}}{0.010 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=118.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

K = 23.15 × 10^-2 S m^-1, c =0.020 M

Then, k = 23.15 × 10^-4 S cm^-1, cm^1/2 = 0. 1414 M^1/2

⇒ \(\Delta_m=\frac{K}{c}=\frac{23.15 \times 10^{-4} \mathrm{Scm}^{-1}}{0.020 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=115.8 \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

k = 55.53 × 10^-2 Sm^-1 ,c = 0.050 M

Then, k = 55.53 x 10^-4 S cm^-1, c^1/2 = 0.2236 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa}{\mathrm{c}}=\frac{55.53 \times 10^{-4} \mathrm{Scm}^{-1}}{0.050 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^2}{\mathrm{~L}}=111.11 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

k= 106.74 × 10^-2 Sm^-1, c = 0.100M

Then, k = 106.74 × 10^-4 Sm^-1, c^1/2 = 0.3162 M^1/2

⇒ \(\Delta_m=\frac{K}{c}=\frac{106.74 \times 10^{-4}}{0.100 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=106.74 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Now, we have the following data.

Since the line interrupts Δ_m at 124.0 S cm² mol^-1 , Δ^o_m = 1 24.0 S cm² mol^-1

Question 11. The conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if Δ^o_m for acetic acid is 390.5 S cm2 mol^-1, what is its dissociation constant?
Answer:

Given, K = 7.896 x 10^-5 S cm^-1

C = 0.00241 mol L^-1

Then, molar conductivity,

⇒ \(\Lambda_m=\frac{\kappa}{c}=\frac{7.896 \times 10^{-5} \mathrm{Scm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=32.76 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Again, = 390.5 S cm mol^-1

Now, \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{mi1}}^0}=\frac{32.76 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}{390.5 \mathrm{Scm}^2 \mathrm{~mol}^{-1}}=0.084\)

∴ Dissociation constant,

⇒ \(\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)}=\frac{\left(0.00241 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.084)^2}{(1-0.084)}=1.86 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

Question 12. How much charge is required for the following reductions:

1 mol of Al³⁺ to Al.

1 mol of Cu²⁺ to Cu.

1 mol of MnO^–₄ to Mn²⁺

Answer:

Al³⁺ + 3e^– → Al

∴ Required charge = 3F

= 3 × 96487 C = 289461 C

Cu²⁺+ 2e^– → Cu

∴ Required charge = 2F

= 2 × 96487 C = 1 92974 C

MnO^–₄ → Mn²⁺

i.e., Mn⁷⁺ + 5e^– → Mr²⁺

∴ Required charge = 5F

= 5 × 96487 C = 482435 C

Question 13. How much electricity in terms of Faraday is required to produce

1. 20.0 g of Ca from molten CaCI₂
2. 40.0 g of Al from molten Al₂O₃.

Answer:

According to the question.

⇒ \(\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{40 \mathrm{~g}}{\mathrm{Ca}}\)

Electricity required to produce 40 g of calcium = 2F

Therefore, electricity required to produce 20 g of calcium = \(\frac{2 \times 20}{40} \mathrm{~F}=1 \mathrm{~F}\)

According to the question

⇒ \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{27 \mathrm{~g}}\)

Electricity required to produce 27 g of Al = 3F

Therefore, electricity required to produce 40 g of Al = \(\frac{3 \times 40}{27} \mathrm{~F}=4.44 \mathrm{~F}\)

Question 14. How much electricity is required in coulomb for the oxidation of

1. 1 mol of I H₂O to O₂.
2. 1 mol of FeO to Fe₂,O₃

Answer:

According to the question,

⇒ \(\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2\)

Now, we can write:

⇒ \(\mathrm{O}^{2-} \longrightarrow \frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)

Electricity required for the oxidation of 1 mol of H₂O to O₂ = 2F = 2 × 96487 C = 192974 C

According to the question,

⇒ \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1}\)

Electricity required for the oxidation of 1 mol of FeO to Fe₂O₃, = 1F = 96487 C

Question 15. A solution of Ni. (NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Nt is deposited at the cathode?
Answer:

Given,

Current = 5A

Time = 20×60= 1200 s

∴ Charge = current x time

= 5 x 1200 = 6000 C.

According to the reaction,

⇒ \(\mathrm{Ni}_{(\mathrm{iq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{58.7 g}{\mathrm{Ni}_{(\mathrm{s})}}\)

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C = \(\frac{58.71 \times 6000}{2 \times 96487} \mathrm{~g}=1.825 \mathrm{~g}\)

Hence, 1.825 g of nickel will be deposited at the cathode

Question 16. Three electrolytic cells A, B. C containing solutions of ZnSO₄, AgNO, and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B, How long did the current flow? What mass of copper and zinc were deposited?
Answer:

According to the reaction:

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \underset{108 \mathrm{~g}}{\mathrm{Ag}_{(\mathrm{s})}}\)

i.e., 108 g of Ag is deposited by 96487 C

Therefore, 1 .45 g of Ag will be deposited by = \(\frac{96487 \times 1.45}{108} C=1295.43 \mathrm{C}\)

Given, Current = 1.5 A

⇒ \(\text { Time }=\frac{1295.43}{1.5} \mathrm{~s}=863.6 \mathrm{~s}=864 \mathrm{~s}=14.40 \mathrm{~min}\)

Again,

⇒ \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{63.5 \mathrm{~g}}{\mathrm{Cu}_{(\mathrm{s})}}\)

2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1 295.43 C of charge will deposit = \(\frac{63.5 \times 1295.43}{2 \times 96487} \mathrm{~g}\) = 0.426 g of Cu

Given,

⇒ \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{65.4 \mathrm{~g}}{\mathrm{Zn}_{(\mathrm{s})}}\)

2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1 295.43 C of charge will deposit = \(\frac{65.4 \times 1295.43}{2 \times 96487} \mathrm{~g}=0.439 \mathrm{~g} \text { of } \mathrm{Zn}\)

Question 17. Using the standard electrode potentials given in the Table predict if the reaction between the following is feasible:

1. Fe³⁺(aq) and I^– (aq )
2. Ag³⁺ (aq) and Cu(s)
3. Fe³⁺(aq) and Br'(aq)
4. Ag(s) and Fe³⁺(aq)
5. Br₂(aq) and Fe²⁺(aq).

Answer:

Since E^– for the overall reaction is positive, the reaction between Fe³⁺_(aq), and I^–_(aq) is feasible.

Since E^– for the overall reaction is positive, the reaction between Ag⁺_(aq) and Cu_(s) is feasible.

Since E^– for the overall reaction is negative, the reaction between Fe³⁺_(aq), and Br_(aq) is feasible.

Since E^– E for the overall reaction is negative, the reaction between Ag_(s) and Fe²⁺_(aq) is not feasible.

Since E^– for the overall reaction is positive, the reaction between Br_2(aq) and Fe²⁺_(aq) is feasible.

Question 18. Predict the products of electrolysis in each to the following:

1. An aqueous solution of AgNO₃, with silver electrodes.
2. An aqueous solution of AgNO₃, with platinum electrodes.
3. A dilute solution of H₂SO₄ with platinum electrodes.
4. An aqueous solution of CuCl₂ with platinum electrodes.

Answer:

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \Lambda \mathrm{g}_{(\mathrm{b})} ; \mathrm{E}^{\ominus}=0.80 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of EM takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At anode: Ag Ag⁺ + e^–

The Ag anode is attacked by NO₃^–, ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag⁺

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; \mathrm{E}^{\ominus}=0.80 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of EM takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At anode: Ag →  Ag⁺ + e^–

Since Pt electrodes arc inert, the anode is not attacked by NOT ions. Therefore, OFT or NOT ions can be oxidized at the anode. But OIF ions have a lower discharge potential and get preference and decompose to liberate O2.

⇒ \(2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} ; \mathrm{E}^0=-1.23 \mathrm{~V}\)

At the cathode, the following reduction occurs to produce H₂(g)

⇒ \(\mathrm{H}_{(aq)}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)}\)

At the anode, the following processes are possible.

⇒ \(2 \mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}: \mathrm{E}^{(-)}=+1.23 \mathrm{~V}\)

⇒ \(2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \longrightarrow \mathrm{S}_2 \mathrm{O}_{6(\mathrm{aq})}^{2-}+2 \mathrm{e}^{-} ; \mathrm{E}^{\ominus}=+1.96 \mathrm{~V}\)

For dilute sulphuric acid, reaction (t) is preferred to produce O₂ gas. But for concentrated sulphuric acid, a reaction occurs.

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Cu}_{(\mathrm{ap})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{g})} ; \mathrm{E}^{\ominus}=0.34 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of E^–  takes place at the die cathode. Therefore, the deposition of copper will take place at the cathode:

At anode:

The following oxidation reactions are possible at the anode.

⇒ \(\mathrm{Cl}_{\text {(aq) }}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_{2(8)}+\mathrm{e}: \mathrm{E}^{\ominus}=-1.36 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}_{(1)} \longrightarrow \mathrm{O}_{2(\mathrm{p})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}: \mathrm{E}^{\ominus}=-1.23 \mathrm{~V}\)

At the anode, the reaction with a lower value of E^– is preferred. However due to the over-potential of oxygen, Cl^– gets oxidized at the anode to produce Cl₂ gas.