Electrochemistry — NCERT Solutions

For each solution,

$$\Lambda_m=\frac{\kappa}{c}$$

Since the table gives $10^2\times\kappa$ in $\mathrm{S\,m^{-1}}$, first write

$$\kappa=(\text{given value})\times10^{-2}\,\mathrm{S\,m^{-1}}$$

Then

$$\Lambda_m=\frac{\kappa}{c}$$

in $\mathrm{S\,m^2\,mol^{-1}}$, and multiply by $10^4$ to get $\mathrm{S\,cm^2\,mol^{-1}}$.

Calculations

1) $c=0.001\,\mathrm M$

$$\kappa=1.237\times10^{-2}=0.01237\,\mathrm{S\,m^{-1}}$$

$$\Lambda_m=\frac{0.01237}{1}=0.01237\,\mathrm{S\,m^2\,mol^{-1}}$$

$$\Lambda_m=123.7\,\mathrm{S\,cm^2\,mol^{-1}}$$

2) $c=0.010\,\mathrm M$

$$\kappa=11.85\times10^{-2}=0.1185\,\mathrm{S\,m^{-1}}$$

$$\Lambda_m=\frac{0.1185}{10}=0.01185\,\mathrm{S\,m^2\,mol^{-1}}$$

$$\Lambda_m=118.5\,\mathrm{S\,cm^2\,mol^{-1}}$$

3) $c=0.020\,\mathrm M$

$$\kappa=23.15\times10^{-2}=0.2315\,\mathrm{S\,m^{-1}}$$

$$\Lambda_m=\frac{0.2315}{20}=0.011575\,\mathrm{S\,m^2\,mol^{-1}}$$

$$\Lambda_m=115.75\,\mathrm{S\,cm^2\,mol^{-1}}$$

4) $c=0.050\,\mathrm M$

$$\kappa=55.53\times10^{-2}=0.5553\,\mathrm{S\,m^{-1}}$$

$$\Lambda_m=\frac{0.5553}{50}=0.011106\,\mathrm{S\,m^2\,mol^{-1}}$$

$$\Lambda_m=111.06\,\mathrm{S\,cm^2\,mol^{-1}}$$

5) $c=0.100\,\mathrm M$

$$\kappa=106.74\times10^{-2}=1.0674\,\mathrm{S\,m^{-1}}$$

$$\Lambda_m=\frac{1.0674}{100}=0.010674\,\mathrm{S\,m^2\,mol^{-1}}$$

$$\Lambda_m=106.74\,\mathrm{S\,cm^2\,mol^{-1}}$$

Table

A plot of $\Lambda_m$ versus $c^{1/4}$ is nearly a straight line. Extrapolating to $c^{1/4}=0$ gives the limiting molar conductivity.

From the given data, the intercept is approximately

$$\Lambda_m^0 \approx 126.4\,\mathrm{S\,cm^2\,mol^{-1}}$$

So the limiting molar conductivity of NaCl is about 126.4 $\mathrm{S\,cm^2\,mol^{-1}}$.

Use

$$\Lambda_m = \frac{\kappa \times 1000}{c}$$

Given:

• $\kappa = 7.896\times 10^{-5}\,\text{S cm}^{-1}$
• $c = 0.00241\,\text{mol L}^{-1}$

So,

$$\Lambda_m = \frac{7.896\times 10^{-5}\times 1000}{0.00241}$$
$$= \frac{0.07896}{0.00241} \approx 32.76\,\text{S cm}^2\text{ mol}^{-1}$$

Now,

$$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{32.76}{390.5} \approx 0.0839$$

Dissociation constant:

$$K_a = \frac{c\alpha^2}{1-\alpha}$$

$$K_a = \frac{0.00241\times (0.0839)^2}{1-0.0839} \approx 1.96\times 10^{-5}$$

The textbook's given in-text answer format for a similar acetic acid example uses this method. Here the computed values from the data are:

$\Lambda_m \approx 32.76\,\text{S cm}^2\text{ mol}^{-1}$ and $K_a \approx 1.96\times 10^{-5}$.

Use Faraday's law: the charge needed equals the number of electrons required per mole of substance multiplied by Faraday, $F$.

(i) $1$ mol of $\mathrm{Al^{3+}}$ to Al

Reduction:

$$\mathrm{Al^{3+}}+3e^-\rightarrow \mathrm{Al}$$

So, charge required = $3F$.

(ii) $1$ mol of $\mathrm{Cu^{2+}}$ to Cu

Reduction:

$$\mathrm{Cu^{2+}}+2e^-\rightarrow \mathrm{Cu}$$

So, charge required = $2F$.

(iii) $1$ mol of $\mathrm{MnO_4^-}$ to $\mathrm{Mn^{2+}}$

In acidic medium:

$$\mathrm{MnO_4^-}+8\mathrm H^++5e^-\rightarrow \mathrm{Mn^{2+}}+4\mathrm H_2\mathrm O$$

So, charge required = $5F$.

Therefore the required charges are:

• (i) $3F$
• (ii) $2F$
• (iii) $5F$

Use atomic masses and the number of electrons required.

(i) $20.0\,\mathrm g$ of Ca from molten $\mathrm{CaCl_2}$

Reaction:

$$\mathrm{Ca^{2+}}+2e^-\rightarrow \mathrm{Ca}$$

1 mol Ca = 40 g requires 2 F.

For 20.0 g Ca:

$$\frac{20.0}{40.0}\times 2F = 1F$$

So electricity required = 1 Faraday.

(ii) $40.0\,\mathrm g$ of Al from molten $\mathrm{Al_2O_3}$

Reaction:

$$\mathrm{Al^{3+}}+3e^-\rightarrow \mathrm{Al}$$

1 mol Al = 27 g requires 3 F.

For 40.0 g Al:

$$\frac{40.0}{27.0}\times 3F \approx 4.44F$$

So electricity required = $4.44F$.

Thus,

• (i) 1 F
• (ii) 4.44 F

(i) 1 mol of $\mathrm{H_2O}$ to $\mathrm{O_2}$

Oxidation reaction:

$$2\mathrm H_2\mathrm O \rightarrow \mathrm O_2 + 4\mathrm H^+ + 4e^-$$

So, 2 mol of water require 4 mol electrons.

Therefore 1 mol of water requires 2 mol electrons = 2F.

Charge in coulombs:

$$2F = 2\times 96487 = 1.92974\times10^5\,\mathrm C$$

(ii) 1 mol of FeO to $\mathrm{Fe_2O_3}$

In FeO, iron is in the +2 state and in $\mathrm{Fe_2O_3}$ it is +3, so each Fe atom loses 1 electron.

To convert 2 mol FeO to 1 mol $\mathrm{Fe_2O_3}$, 2 mol electrons are needed. Hence 1 mol FeO needs 1 mol electron = 1F.

Charge in coulombs:

$$1F = 96487\,\mathrm C$$

Therefore,

• (i) $1.93\times10^5\,\mathrm C$
• (ii) $9.65\times10^4\,\mathrm C$

Use Faraday's law.

Given:

• Current, $I=5$ A
• Time, $t=20$ min $=1200$ s

Charge passed:

$$Q=It=5\times 1200=6000\,\text{C}$$

For nickel deposition:

$$\mathrm{Ni^{2+} + 2e^- \rightarrow Ni}$$

So 2F deposits 58.7 g Ni.

Mass deposited:

$$m=\frac{Q}{2F}\times 58.7$$

$$m=\frac{6000}{2\times 96487}\times 58.7$$

$$m\approx 1.83\,\text{g}$$

So the deposited mass is 1.83 g.

Silver deposited at cell B is 1.45 g.

For silver:
$$\mathrm{Ag^+ + e^- \rightarrow Ag}$$
So 1 mol Ag needs 1F and has mass 108 g.

Charge used:

$$Q=\frac{1.45}{108}\times 96487$$
$$Q\approx 1296\,\text{C}$$

Current = 1.5 A, so time:

$$t=\frac{Q}{I}=\frac{1296}{1.5}=864\,\text{s}$$

$$864\,\text{s}=14.4\,\text{min}$$

For copper deposition:
$$\mathrm{Cu^{2+}+2e^- \rightarrow Cu}$$
Mass deposited by same charge:

$$m_{Cu}=\frac{Q}{2F}\times 63.5$$
$$=\frac{1296}{2\times 96487}\times 63.5 \approx 0.427\,\text{g}$$

For zinc deposition:
$$\mathrm{Zn^{2+}+2e^- \rightarrow Zn}$$
Mass deposited:

$$m_{Zn}=\frac{Q}{2F}\times 65.4 \approx 0.439\,\text{g}$$

Thus the correct values are 14.4 min, 0.427 g Cu, 0.439 g Zn.

Use the standard electrode potentials from the table. A reaction is feasible if the cell potential is positive.

Relevant values:

• $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=0.77\,\mathrm V$
• $E^\circ(\mathrm{I_2/I^-})=0.54\,\mathrm V$
• $E^\circ(\mathrm{Ag^+/Ag})=0.80\,\mathrm V$
• $E^\circ(\mathrm{Cu^{2+}/Cu})=0.34\,\mathrm V$
• $E^\circ(\mathrm{Br_2/Br^-})=1.09\,\mathrm V$

(i) $\mathrm{Fe^{3+}}$ and $\mathrm{I^-}$

$\mathrm{Fe^{3+}}$ is reduced and $\mathrm{I^-}$ is oxidised:

$$E^\circ_{cell}=0.77-0.54=0.23\,\mathrm V$$

Positive, so feasible.

(ii) $\mathrm{Ag^+}$ and $\mathrm{Cu(s)}$

$$E^\circ_{cell}=0.80-0.34=0.46\,\mathrm V$$

Positive, so feasible.

(iii) $\mathrm{Fe^{3+}}$ and $\mathrm{Br^-}$

$$E^\circ_{cell}=0.77-1.09=-0.32\,\mathrm V$$

Negative, so not feasible.

(iv) $\mathrm{Ag(s)}$ and $\mathrm{Fe^{3+}}$

Silver metal would be oxidised by $\mathrm{Fe^{3+}}$ only if the cell potential were positive. The reaction would be:

$$\mathrm{Fe^{3+}}+\mathrm{Ag}\rightarrow \mathrm{Fe^{2+}}+\mathrm{Ag^+}$$

$$E^\circ_{cell}=0.77-0.80=-0.03\,\mathrm V$$

Negative, so not feasible.

(v) $\mathrm{Br_2}$ and $\mathrm{Fe^{2+}}$

$$E^\circ_{cell}=1.09-0.77=0.32\,\mathrm V$$

Positive, so feasible.

Hence the feasible reactions are (i), (ii) and (v); (iii) and (iv) are not feasible.

Predict products by comparing the possible electrode reactions.

(i) Aqueous $\mathrm{AgNO_3}$ with silver electrodes

With silver electrodes, the anode itself dissolves and the cathode gets silver deposited.

• Cathode: $\mathrm{Ag^+ + e^- \rightarrow Ag}$
• Anode: $\mathrm{Ag \rightarrow Ag^+ + e^-}$

So the electrolyte composition remains essentially unchanged.

(ii) Aqueous $\mathrm{AgNO_3}$ with platinum electrodes

Platinum is inert.

• Cathode: $\mathrm{Ag^+}$ is reduced to Ag(s)
• Anode: water is oxidised to $\mathrm{O_2}$

So products are silver at cathode and oxygen at anode.

(iii) Dilute $\mathrm{H_2SO_4}$ with platinum electrodes

Platinum is inert.

• Cathode: $\mathrm{H^+}$ is reduced to $\mathrm{H_2}$
• Anode: water is oxidised to $\mathrm{O_2}$

So products are hydrogen at cathode and oxygen at anode.

(iv) Aqueous $\mathrm{CuCl_2}$ with platinum electrodes

With inert Pt electrodes:

• Cathode: $\mathrm{Cu^{2+}}$ is reduced to Cu(s)
• Anode: $\mathrm{Cl^-}$ is oxidised to $\mathrm{Cl_2}$

So products are copper at cathode and chlorine at anode.

Therefore:

• (i) Ag deposited at cathode; Ag dissolves at anode
• (ii) Ag at cathode; O$_2$ at anode
• (iii) H$_2$ at cathode; O$_2$ at anode
• (iv) Cu at cathode; Cl$_2$ at anode