Electrochemistry

A metal higher in the reactivity series displaces a metal lower than itself from its salt solution. So among these metals, the decreasing order of reactivity is Mg, Al, Zn, Fe, Cu. Hence the order in which they displace each other is:

Mg > Al > Zn > Fe > Cu.

Reducing power increases as the standard reduction potential becomes more negative. From the given values:
- Ag⁺/Ag = +0.80 V
- Hg²⁺/Hg = +0.79 V
- Cr³⁺/Cr = -0.74 V
- Mg²⁺/Mg = -2.37 V
- K⁺/K = -2.93 V

So the increasing order of reducing power is:

Ag < Hg < Cr < Mg < K.

The galvanic cell is:

Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

- Negatively charged electrode: Zn electrode because oxidation takes place there, so it is the anode.
- Current carriers in the cell:
- In the external circuit, electrons flow from Zn to Ag.
- In the cell, the ions in the electrolytes and salt bridge carry charge.
- Half-reactions:
- Anode (oxidation): $\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$
- Cathode (reduction): $\mathrm{2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)}$

Thus the cell converts the chemical energy of the spontaneous reaction into electrical energy.

The standard cell potential is found from

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$$

### (i) $2\mathrm{Cr(s)} + 3\mathrm{Cd}^{2+}(aq) \rightarrow 2\mathrm{Cr}^{3+}(aq) + 3\mathrm{Cd(s)}$

Given standard reduction potentials:
- $\mathrm{Cr}^{3+}/\mathrm{Cr} = -0.74\,\mathrm{V}$
- $\mathrm{Cd}^{2+}/\mathrm{Cd} = -0.40\,\mathrm{V}$

Here Cd²⁺ is reduced and Cr is oxidised.

$$E^\circ_{\text{cell}} = (-0.40)-(-0.74)=+0.34\,\text{V}$$

Now,
$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$$

Here $n=6$ electrons.

$$\Delta_r G^\circ = -6\times 96487\times 0.34$$
$$\Delta_r G^\circ \approx -1.97\times 10^5\,\text{J mol}^{-1} = -197\,\text{kJ mol}^{-1}$$

For equilibrium constant:

$$\Delta_r G^\circ = -RT\ln K$$

or at 298 K,

$$\log K = \frac{nE^\circ_{\text{cell}}}{0.059} = \frac{6\times 0.34}{0.059} \approx 34.6$$

$$K \approx 4.0\times 10^{34}$$

### (ii) $\mathrm{Fe}^{2+}(aq)+\mathrm{Ag}^+(aq)\rightarrow \mathrm{Fe}^{3+}(aq)+\mathrm{Ag(s)}$

Relevant potentials:
- $\mathrm{Ag^+}/\mathrm{Ag} = +0.80\,\mathrm{V}$
- $\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}} = +0.77\,\mathrm{V}$

So,

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

Here $n=1$.

$$\Delta_r G^\circ = -1\times 96487\times 0.03 \approx -2.89\times 10^3\,\text{J mol}^{-1} = -2.9\,\text{kJ mol}^{-1}$$

And,

$$\log K = \frac{1\times 0.03}{0.059} \approx 0.51$$

$$K \approx 3.2$$

Use the Nernst equation at 298 K:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q$$

### (i) $\mathrm{Mg(s)}|\mathrm{Mg}^{2+}(0.001\,M)||\mathrm{Cu}^{2+}(0.0001\,M)|\mathrm{Cu(s)}$

Overall reaction:

$$\mathrm{Mg(s)} + \mathrm{Cu}^{2+}(aq) \rightarrow \mathrm{Mg}^{2+}(aq) + \mathrm{Cu(s)}$$

Here, $n=2$ and

$$Q=\frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Cu}^{2+}]}=\frac{0.001}{0.0001}=10$$

Given from the chapter:
$$E^\circ_{\text{cell}} = 0.34 - (-2.37)=2.71\,\text{V}$$

So,

$$E_{\text{cell}} = 2.71 - \frac{0.059}{2}\log 10$$
$$=2.71-0.0295\approx 2.68\,\text{V}$$

### (ii) $\mathrm{Fe(s)}|\mathrm{Fe}^{2+}(0.001\,M)||\mathrm{H}^{+}(1\,M)|\mathrm{H}_2(1\,bar)|\mathrm{Pt(s)}$

The cell reaction is:

$$\mathrm{Fe(s)} + 2\mathrm{H}^+(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{H}_2(g)$$

Here, $n=2$ and

$$Q=\frac{[\mathrm{Fe}^{2+}]\,P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}=\frac{0.001\times 1}{1^2}=0.001$$

Standard emf:

$$E^\circ_{\text{cell}} = 0.00 - (-0.44)=0.44\,\text{V}$$

Thus,

$$E_{\text{cell}} = 0.44 - \frac{0.059}{2}\log(0.001)$$
$$=0.44 - 0.0295(-3)=0.44+0.0885=0.53\,\text{V}$$

### (iii) $\mathrm{Sn(s)}|\mathrm{Sn}^{2+}(0.050\,M)||\mathrm{H}^+(0.020\,M)|\mathrm{H}_2(1\,bar)|\mathrm{Pt(s)}$

Cell reaction:

$$\mathrm{Sn(s)} + 2\mathrm{H}^+(aq) \rightarrow \mathrm{Sn}^{2+}(aq) + \mathrm{H}_2(g)$$

Here, $n=2$ and

$$Q=\frac{[\mathrm{Sn}^{2+}]\,P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}=\frac{0.050}{(0.020)^2}=125$$

Standard emf:

$$E^\circ_{\text{cell}} = 0.00 - (-0.14)=0.14\,\text{V}$$

So,

$$E_{\text{cell}} = 0.14 - \frac{0.059}{2}\log 125$$
$$\log 125 \approx 2.097$$

$$E_{\text{cell}} = 0.14 - 0.0295(2.097) \approx 0.08\,\text{V}$$

### (iv) $\mathrm{Pt(s)}|\mathrm{Br}^-(0.010\,M)|\mathrm{Br}_2(l)||\mathrm{H}^+(0.030\,M)|\mathrm{H}_2(1\,bar)|\mathrm{Pt(s)}$

The relevant half-cells are:
- $\mathrm{Br}_2 + 2e^- \rightarrow 2\mathrm{Br}^-$, $E^\circ=1.09\,\text{V}$
- $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2$, $E^\circ=0.00\,\text{V}$

So the cathode is bromine and the anode is hydrogen.

Cell reaction:

$$\mathrm{Br}_2(l) + \mathrm{H}_2(g) \rightarrow 2\mathrm{Br}^-(aq) + 2\mathrm{H}^+(aq)$$

For this reaction,

$$Q=[\mathrm{Br}^-]^2[\mathrm{H}^+]^2 = (0.010)^2(0.030)^2$$

But since the problem asks for emf, we can use the half-cell form directly:

$$E^\circ_{\text{cell}} = 1.09 - 0.00 = 1.09\,\text{V}$$

and the Nernst equation gives the emf. At 298 K,

$$E_{\text{cell}} = 1.09 - \frac{0.059}{2}\log\left([\mathrm{Br}^-]^2[\mathrm{H}^+]^2\right)$$

$$=1.09 - 0.059\log\left([\mathrm{Br}^-][\mathrm{H}^+]\right)$$

Substituting values:

$$[\mathrm{Br}^-][\mathrm{H}^+] = 0.010\times 0.030 = 3.0\times 10^{-4}$$

$$\log(3.0\times 10^{-4}) = -3.523$$

$$E_{\text{cell}} = 1.09 - 0.059(-3.523) \approx 1.30\,\text{V}$$

For the button cell reaction,

$$\mathrm{Zn(s)} + \mathrm{Ag_2O(s)} + \mathrm{H_2O(l)} \rightarrow \mathrm{Zn}^{2+}(aq) + 2\mathrm{Ag(s)} + 2\mathrm{OH}^-(aq)$$

From the chapter, this reaction involves transfer of 2 electrons, so $n=2$.

The standard cell potential for this cell is obtained from standard reduction potentials:
- $\mathrm{Ag_2O/Ag}$ has the higher reduction potential
- $\mathrm{Zn^{2+}/Zn} = -0.76\,\text{V}$

Using the standard result for the button cell, the emf is about

$$E^\circ = 1.56\,\text{V}$$

Now,

$$\Delta_r G^\circ = -nFE^\circ$$

$$\Delta_r G^\circ = -2\times 96487\times 1.56$$

$$\Delta_r G^\circ \approx -3.01\times 10^5\,\text{J mol}^{-1}$$

$$\Delta_r G^\circ \approx -301\,\text{kJ mol}^{-1}$$

Conductivity $\kappa$ is the conductance of a solution of unit length and unit cross-sectional area. It is the inverse of resistivity. Its SI unit is **S m$^{-1}$.

Molar conductivity** $\Lambda_m$ is the conductivity of the volume of solution containing 1 mole of electrolyte. It is given by

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

where $c$ is the concentration.

### Variation with concentration
- Conductivity decreases on dilution because the number of ions per unit volume decreases.
- Molar conductivity increases on dilution because the volume containing 1 mole of electrolyte increases more than the decrease in conductivity.
- For strong electrolytes, $\Lambda_m$ increases slowly with dilution.
- For weak electrolytes, $\Lambda_m$ increases sharply on dilution because the degree of dissociation increases.

At infinite dilution, molar conductivity is called limiting molar conductivity $\Lambda_m^\circ$.

Use

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

Given:
- $\kappa = 0.0248\,\text{S cm}^{-1}$
- $c = 0.20\,\text{mol L}^{-1}$

For molar conductivity in $\text{S cm}^2\text{ mol}^{-1}$,

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

$$\Lambda_m = \frac{0.0248 \times 1000}{0.20}$$
$$= \frac{24.8}{0.20} = 124\,\text{S cm}^2\text{ mol}^{-1}$$

So the correct value is 124 S cm² mol⁻¹. If the value is compared with the book, it is the same computation used there.

Cell constant is

$$G^* = \kappa R$$

Given:
- $\kappa = 0.146\times 10^{-3}\,\text{S cm}^{-1}$
- $R = 1500\,\Omega$

So,

$$G^* = 0.146\times 10^{-3} \times 1500$$
$$= 0.219\,\text{cm}^{-1}$$

But the standard result for the same data in the textbook exercise uses the conductivity value as given and the cell constant comes out by direct multiplication. If computed strictly from the given numbers, the cell constant is 0.219 cm⁻¹.