Answer

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

$\mathrm{Mg}, \mathrm{Al}, \mathrm{Zn}, \mathrm{Fe}, \mathrm{Cu}$

Answer

Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be $\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K}$ .

Answer

The set-up will be similar to that . The cell will be represented as :

$ \mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s) $

(i) Anode, i.e., zinc electrode will be negatively charged.

(ii) The current will flow from silver to copper in the external circuit.

(iii) At Anode : $\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-}$

At Cathode : $\mathrm{Ag}^{+}(a q)+e \longrightarrow \mathrm{Ag}$

Answer

(i) $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=0.74 \mathrm{~V}$

$E^{o}{ _{\mathrm{Cd}^{2+} / \mathrm{Cd}}}=-0.40 \mathrm{~V}$

The galvanic cell of the given reaction is depicted as :

$\mathrm{Cr {(s)}} |\mathrm{Cr^{3+} {(a q)}}||\mathrm{Cd^{2+} {(a q)}}| \mathrm{Cd {(s)}}$

Now, the standard cell potential is : $ E_{\text {cell }}^{o} =E_{{R}}^{o}-E_{{L}}^{o} $

$ \hspace{7cm} =-0.40-(-0.74) $

$ \hspace{7cm} =+0.34 {~V} $

In the given equation, $n=6$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{F}=96487 \hspace{0.5mm} \mathrm{C} \hspace{0.5mm} \mathrm{mol}^{-1}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad E_{\text {cell }}^{o}=+0.34 \mathrm{~V}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\Delta_{{r}} G^{o} =-n {~F} E_{\text {cell }}^{o}$

Then, $\Delta_{\mathrm{r}} G^{o}=-6 \times 96487 \hspace{0.5mm} \mathrm{C} \hspace{0.5mm} \mathrm{mol}^{-1} \times 0.34 \mathrm{~V}$

$ \hspace{2.2cm} =-196833.48 \hspace{0.5mm}\mathrm{CV}\hspace{0.5mm} \mathrm{mol}^{-1}$

$ \hspace{2.2cm} =-196833.48 \mathrm{~J} \mathrm{~mol}^{-1}$

$ \hspace{2.2cm} =-196.83 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Again, $\Delta_{\mathrm{r}} G^{o}=-\mathrm{R} T \ln K$

$\Rightarrow \hspace{2.5mm} \Delta_{\mathrm{r}} G^{o}=-2.303 \mathrm{R} T \ln K$

$\Rightarrow \hspace{2.5mm} \log K=-\dfrac{\Delta_{\mathrm{r}} G}{2.303 \mathrm{R} T}$

$ \hspace{1.9cm} =\dfrac{-196.83 \times 10^{3}}{2.303 \times 8.314 \times 298} $

$\hspace{1.9cm}=34.496$

$\therefore \mathrm{K}= \text{antilog (34.496)}$

$ \hspace{0.9cm}=3.13 \times 10^{34}$

(ii) $E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{o}=0.77 \mathrm{~V}$

$E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{o}=0.80 \mathrm{~V}$

The galvanic cell of the given reaction is depicted as:

$ \mathrm{Fe^{2+}{(a q)}} \left|\mathrm{Fe^{3+}{(a q)}}\right|\left|\mathrm{Ag^{+}{(a q)}}\right| \mathrm{Ag{(s)}} $

Now, the standard cell potential is $E_{\text {cell }}^{o} =E_{\mathrm{R}}^{o}-E_{\mathrm{L}}^{o} $

$ \hspace{6.8cm} =0.80-0.77 $

$ \hspace{6.8cm} =0.03 \mathrm{~V}$

$ Here, n=1$.

$ Then, \Delta_{\mathrm{r}} G^{o}=-n \mathrm{~F} E_{\text {cell }}^{o}$

$ \hspace{2.3cm} =-1 \times 96487 \hspace{0.5mm} \mathrm{C} \hspace{0.5mm} \mathrm{mol}^{-1} \times 0.03 \mathrm{~V}$

$\hspace{2.3cm}=-2894.61 \hspace{0.5mm}\mathrm{CV}\hspace{0.5mm} \mathrm{mol}^{-1}$

$\hspace{2.3cm}=-2894.61 \mathrm{~J} \mathrm{~mol}^{-1}$

$\hspace{2.3cm}=-2.89 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\text { Again, } \Delta_{\mathrm{r}} G^{o}=-2.303 \mathrm{R} T \ln K $

$\Rightarrow \log K=-\dfrac{\Delta_{\mathrm{r}} G}{2.303 \mathrm{R} T}$

$ \hspace{1.6cm} =\dfrac{-2894.61}{2.303 \times 8.314 \times 298}$

$ \hspace{1.6cm}=0.5073$

$\therefore \mathrm{K}=\text { antilog (0.5073) } $

$\quad\quad=3.2 \text { (approximately) }$

Answer

(i) Cell reaction : $\mathrm{Mg}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Cu}(n=2)$

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

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad E_{\text {cell }} =E_{\text {cell }}^{o}-\dfrac{0.0591}{n} \log \dfrac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad={0.34-(-2.36)}-\dfrac{0.0591}{2} \log \dfrac{0.001}{0.0001} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=2.7-\dfrac{0.0591}{2} \log 10 $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=2.7-0.02955$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=2.67 \mathrm{~V}$ (approximately)

(ii) Cell reaction : $\mathrm{Fe}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{H}_2(n=2)$

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

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
E_{\text {cell }}=E_{\text {cell }}^{o}-\dfrac{0.0591}{n} \log \dfrac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} $

$ \hspace{6.2cm} ={0-(-0.44)}-\dfrac{0.0591}{2} \log \dfrac{0.001}{1^{2}} $

$ \hspace{6.2cm} =0.44-0.02955(-3) $

$ \hspace{6.2cm} = 0.52865 \mathrm{~V}$

$ \hspace{6.2cm} =0.53 \mathrm{~V}$ (approximately)

(iii) $\text { Cell reaction : } \mathrm{Sn}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{H}_2(n=2) $

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

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad E_{\text {cell }} =E_{\text {cell }}^{o}-\dfrac{0.0591}{n} \log \dfrac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad ={0-(-0.14)}-\dfrac{0.0591}{2} \log \dfrac{0.050}{(0.020)^{2}}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =0.14-0.0295 \times \log 125$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =0.14-0.062$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =0.078 \mathrm{~V}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =0.08 \mathrm{~V}$ (approximately)

(iv)$\text { Cell reaction : } 2 \mathrm{Br}^{-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Br}_2+\mathrm{H}_2 \text { (n=2) }$

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

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad E_{\text {cell }} =E_{\text {cell }}^{o}-\dfrac{0.0591}{n} \log \dfrac{1}{\left[\mathrm{Br}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{2}} $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =(0-1.09)-\dfrac{0.0591}{2} \log \dfrac{1}{(0.010)^{2}(0.030)^{2}} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.09-0.02955 \times \log \dfrac{1}{0.00000009} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.09-0.02955 \times \log \dfrac{1}{9 \times 10^{-8}} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.09-0.02955 \times \log \left(1.11 \times 10^{7}\right) $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.09-0.02955(0.0453+7) $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.09-0.208 $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =-1.298 \mathrm{~V}$

Answer

$ \begin{array}{rl} \mathrm{Zn\ (s)} & \longrightarrow \mathrm{Zn^{2+}\ (aq)} + 2 \mathrm{e}^{-} ; E^{o} = 0.76\ \mathrm{V} \ \mathrm{Ag_2O\ (s)} + \mathrm{H_2O\ (l)} + 2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Ag\ (s)} + 2 \mathrm{OH^{-}\ (aq)} ; E^{o} = 0.344\ \mathrm{V} \ \hline \mathrm{Zn\ (s)} + \mathrm{Ag_2O\ (s)} + \mathrm{H_2O\ (l)} & \longrightarrow \mathrm{Zn^{2+}\ (aq)} + 2 \mathrm{Ag\ (s)} + 2 \mathrm{OH^{-}\ (aq)} ; E^{o} = 1.104\ \mathrm{V} \end{array} $

$\therefore E^{o}=1.104 \mathrm{~V}$

We know that, $\Delta_{r} G^{o}=-n \mathrm{~F} E^{o} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad = -2 \times 96487 \times 1.104 $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad = -213043.296 \mathrm{~J} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad = -213.04 \mathrm{~kJ}$

Answer

Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross section.The conductivity is also known as specific conductivity.

$ \text { Specific conductivity }(\kappa)=\dfrac{1}{\rho}=\dfrac{1}{\text { ohm cm }}=\mathrm{ohm}^{-1} \mathrm{~cm}^{-1} $

Molar conductivity:

Molar conductivity of a solution at a dilution $V$ is the conductance of all the ions produced from one mole of the electrolyte dissolved in $V \mathrm{~cm}^3$ of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by $\Lambda^c_m$.

$\Lambda^c_m=\kappa \times {V}$

$ \Lambda^c_m=\kappa_c \times \dfrac{1000}{c}=\kappa_c \times \dfrac{1000}{\text { Molarity }} $

where $\kappa$ is the specific conductivity and V is the volume of the solution containing one mole of the electrolyte and $c$ is the molar concentration, i.e., $\mathrm{mol}\hspace{0.5mm} \mathrm{L}^{-1}\left(\mathrm{or}\hspace{0.5mm} \mathrm{mol} \hspace{0.5mm}\mathrm{dm}^{-3}\right)$.

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.It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Lm can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte.

The variation of $\Lambda_{m}$ with $\sqrt{c}$ for strong electrolytes (potassium chloride) and weak electrolytes (acetic acid) is shown in the following plot:

Answer

Given,

$\kappa=0.0248 \mathrm{~S} \mathrm{~cm}^{-1}$

$\mathrm{C}=0.20 \hspace{0.5mm}\mathrm{M}$

$\therefore$ Molar conductivity ; $ \Lambda^c_{m}=\dfrac{\kappa \times 1000}{\mathrm{c}} $

$ \hspace{4.6cm} =\dfrac{0.0248 \times 1000}{0.2}$

$ \hspace{4.6cm} =124 \hspace{0.5mm}\mathrm{\hspace{0.5mm}S\hspace{0.5mm}cm}^{2} \mathrm{~mol}^{-1}$

Answer

$\text{Cell constant} =\dfrac{\text { Conductivity }}{\text { Conductance }}$

$\quad\quad\quad\quad\quad\quad=\text { Conductivity } \times \text { Resistance }$

$\quad\quad\quad\quad\quad\quad= \left(0.146 \times 10^{-3}\right) \mathrm{S} \hspace{0.7 mm}\mathrm{cm}^{-1} \times 1500\hspace{1 mm}\Omega $

$\quad\quad\quad\quad\quad\quad= \mathbf{0 . 2 1 9} \hspace{1 mm}\mathrm{cm}^{-1}$

Answer

$\begin{aligned} 1 \mathrm{~S} \mathrm{~cm}^{-1} & =100 \mathrm{~S} \mathrm{~m}^{-1} \ \dfrac{1 \mathrm{~S} \mathrm{~cm}^{-1}}{100 \mathrm{~S} \mathrm{~m}^{-1}} & =1 \text { (unit conversion factor) }\end{aligned}$

$\Lambda^{o}=$ Intercept on the $\Lambda_{\mathrm{m}}$ axis $=124.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ (on extrapotation to zero concentration)

Answer

Given, $\kappa=7.896 \times 10^{-5} \mathrm{~S} \mathrm{~m}^{-1}$

$\mathrm{C}=0.00241 \mathrm{~mol} \mathrm{~L}^{-1}$

Then, molar conductivity, $\Lambda_{m}=\dfrac{\kappa}{\mathrm{c}}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\dfrac{7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}} \times \dfrac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=32.76 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Again, $\Lambda_{m}^{0}=390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Now, $\alpha=\dfrac{\Lambda^c_{m}}{\Lambda_{m}^{0}}=\dfrac{32.76 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}$

$\quad\quad\quad\quad\quad\quad=0.084$

$\therefore$ Dissociation constant, $K_{a}=\dfrac{\mathrm{c} \alpha^{2}}{(1-\alpha)}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\dfrac{\left(0.00241 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.084)^{2}}{(1-0.084)}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=1.86 \times 10^{-5} $

Answer

(i) $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$

$\therefore \text { Quantity of charge required for reduction of } 1 \mathrm{~mol} \text { of } \mathrm{Al}^{3+}=3 \mathrm{~F}=3 \times 96500 \mathrm{~C}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad = {2 8 9 5 0 0} {~C} \text {. }$

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

$\therefore$ Quantity of charge required for reduction of $1 \mathrm{~mol}^{\text {of }} \mathrm{Cu}^{2+}=2$ faradays $=2 \times 96500 \hspace{1mm} \mathrm{C}$

$ \hspace{12.7cm} =193000 \hspace{1mm} \mathrm{C}$

(iii) $\mathrm{MnO_4}^{-} \longrightarrow \mathrm{Mn}^{2+}$

$\therefore$ Quantity of charge required $=5 \mathrm{~F}=5 \times 96500 \hspace{1mm} \mathrm{C}$

$\hspace{6.3cm}=482500 \hspace{1mm} \mathrm{C}$.

Answer

(i) $\mathrm{Ca}^{2+}+2 e^{-} \longrightarrow \mathrm{Ca}$

Thus, 1 mol of Ca , i.e., 40 g of Ca require electricity $=2 \mathrm{~F} $

$\therefore 20 \mathrm{~g}$ of Ca will require electricity $=1 \mathrm{~F}$

(ii) $\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}$.

Thus, 1 mol of Al , i.e., 27 g of Al require electricity $=3 \mathrm{~F}$

$\therefore 40 \mathrm{~g}$ of Al will require electricity $=\dfrac{3}{27} \times 40=4 \cdot 44 \mathrm{~F}$.

Answer

(i) The electrode reaction for 1 mol of $\mathrm{H}_2 \mathrm{O}$ is :

$ \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\dfrac{1}{2} \mathrm{O}_2 \text {, i.e., } \mathrm{O}^{2-} \longrightarrow \dfrac{1}{2} \mathrm{O}_2+2 e^{-} $

$ \text { or } \quad 2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2 $

$\text { or } \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{H}^{+}+\dfrac{1}{2} \mathrm{O}_2+2 e^{-}$

$\therefore$ Quantity of electricity required for oxidation of 1 mol of $\mathrm{H}_2 \mathrm{O}=2 \mathrm{~F}$

$ \hspace{11cm} =2 \times 96500 \hspace{0.5mm}\mathrm{C}$

$ \hspace{11cm} =193000 \hspace{0.5mm} \mathrm{C}$

(ii) The electrode reaction for 1 mol of FeO is :

$\mathrm{FeO} \longrightarrow \dfrac{1}{2} \mathrm{Fe}_2 \mathrm{O}_3$

$ i.e., \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}$

$\therefore$ Quantity of electricity required $=1 \mathrm{~F}=96500 \mathrm{C}$.

Answer

Given,

Current $=5 \mathrm{~A}$

Time $=20 \times 60=1200 \mathrm{~s}$

$\therefore$ Charge $=$ current $\times$ time

$\quad\quad\quad\quad=5 \times 1200$

$\quad\quad\quad\quad=6000 \hspace{0.5mm}\mathrm{C}$

According to the reaction,

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

Thus, $2 F$ , i.e., $2 \times 96500 \mathrm{C}$ deposit $\mathrm{Ni}=1$ mole, i.e., 58.7 g

$ \therefore 6000 \mathrm{C} \text { will deposit } \mathrm{Ni}=\dfrac{58.7}{2 \times 96500} \times 6000 \mathrm{~g}=1.825 \mathrm{~g} $

Hence, $1.825 \mathrm{~g}$ of nickel will be deposited at the cathode.

Answer

According to the reaction:

$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow {\mathrm{Ag}}$

i.e., $108 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $96487 \mathrm{~C}$.

Therefore, $1.45 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $=\dfrac{96487 \times 1.45}{108} \mathrm{~C}$

$ \hspace{6.8cm} =1295.43 \mathrm{~C}$

Given, Current $=1.5 \mathrm{~A}$

$\therefore$ Time $=\dfrac{1295.43}{1.5} \mathrm{~s}=863.6 \mathrm{~s}$

$ \hspace{3.8cm} =864 \mathrm{~s}$

$ \hspace{3.8cm} =14.40 \mathrm{~min}$

Again,

$ \mathrm{Cu_{(\alpha q)}^{2+}}+2 \mathrm{e}^{-} \longrightarrow \underset{63.5 \mathrm{~g}}{\mathrm{Cu_{(s)}}} $

i.e., $2 \times 96487 \mathrm{C}$ of charge deposit $=63.5 \mathrm{~g}$ of $\mathrm{Cu}$

Therefore,$ 1295.43 ~C $ of charge will deposit $ =\dfrac{63.5 \times 1295.43}{2 \times 96487} \mathrm{~g} $

$ \hspace{7.5cm} =0.426 \mathrm{~g}$ of $\mathrm{Cu}$

$ \text{For, } \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow {\mathrm{Zn}} $

i.e., $2 \times 96487 \mathrm{~C}$ of charge deposit $=65.4 \mathrm{~g}$ of $\mathrm{Zn}$

Therefore, $1295.43 \mathrm{~C}$ of charge will deposit $=\dfrac{65.4 \times 1295.43}{2 \times 96487} \mathrm{~g}$

$ \hspace{7.6cm} =0.439 \mathrm{~g}$ of $\mathrm{Zn}$

Answer

A reaction feasible if EMF of the cell reaction is $+ve$

$ \text { (i) } \mathrm{Fe}^{3+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\frac{1}{2} \mathrm{I}_2 $

$\text { i.e., } \mathrm{Pt}\left|\mathrm{I}_2\right| \mathrm{I}^{-}(a q)| | \mathrm{Fe}^{3+}(a q)\left|\mathrm{Fe}^{2+}(a q)\right| \mathrm{Pt} $

$ \therefore \mathrm{E}{\text {cell }}^{\circ}=\mathrm{E}{\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}}^{\circ}-\mathrm{E}_{1 / 2 \mathrm{I}_2, \mathrm{I}^{-}}^{\circ}$

$\quad\quad\quad=0.77-0.54=0.23 \mathrm{~V} \text { (Feasible) } $

Since $E^{o}$ for the overall reaction is positive, the reaction between $\mathrm{Fe^{3+}} $ and $\mathrm{I^-}$ is feasible.

(ii) $ \mathrm{Ag}^{+}(a q)+\mathrm{Cu} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) $

$\text {, i.e., }{\mathrm{Cu}\left|\mathrm{Cu}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}} $

${\mathrm{E}{\text {cell }}^{\circ}=\mathrm{E}{\mathrm{Ag}^{+}, \mathrm{Ag}^{-}}-\mathrm{E}_{\mathrm{Cu}^{2+}, \mathrm{Cu}}}$

$\quad\quad=0.80-0.34=0.46 \mathrm{~V} \text { (Feasible) } .$

Since $E^{\text {o }}$ for the overall reaction is positive, the reaction between $\mathrm{Ag_{(a q)}^{+}}$ and $\mathrm{Cu_{(s)}}$ is feasible.

(iii) $\mathrm{Fe}^{3+}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\frac{1}{2} \mathrm{Br}_2$

$\mathrm{E}_{\text {cell }}^{\circ}=0.77-1.09=-0.32 \mathrm{~V}$ (Not feasible)

Since $E^{0}$ for the overall reaction is negative, the reaction between $\mathrm{Fe^{3+}}$ and $\mathrm{Br^{-}}$ is not feasible.

(iv) $\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q) $

$\mathrm{E}_{\text {cell }}^{\circ}=0.77-0.80=-0.03 \mathrm{~V}$ (Not feasible)

Since $E^{\mathrm{o}} $ for the overall reaction is negative, the reaction between $\mathrm{Ag}$ and $\mathrm{Fe^{3+}}$ is not feasible.

(v) $\frac{1}{2} \mathrm{Br}_2(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Br}^{-}+\mathrm{Fe}^{3+}$

$ \mathrm{E}_{\text {cell }}^{\mathrm{o}}=1.09-0.77=0.32 \mathrm{~V}$ (Feasible)

Since $E^{0}$ for the overall reaction is positive, the reaction between $\mathrm{Br_2(a q)}$ and $\mathrm{Fe}^{2+}{(a q)}$ is feasible.

Answer

(i) Electrolysis of aqueous solution of $\mathrm{AgNO}_3$ with silver electrodes.

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{AgNO}_3(s)+a q \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_3^{-}(a q) $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} $

At Cathode : $\mathrm{Ag}^{+}$ ions have lower discharge potential than $\mathrm{H}^{+}$ ions. Hence, $\mathrm{Ag}^{+}$ ions will be deposited as Ag in preference to $\mathrm{H}^{+}$ ions.

Alternatively, we have standard reduction potentials as :

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s), \mathrm{E}^{\circ}=+0.80 \mathrm{~V}$

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{H}^{+}(a q)+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(g), \mathrm{E}^{\circ}=0.00 \mathrm{~V} $

As $\mathrm{Ag}^{+}$ ions have higher standard reduction potential than that of $\mathrm{H}^{+}$ ions, hence $\mathrm{Ag}^{+}$ ions will be reduced more easily and deposited as Ag.

At Anode : As Ag anode is attacked by $\mathrm{NO}_3^{-}$ ions, Ag of the anode will dissolve to form $\mathrm{Ag}^{+}$ ions in the solution.

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+e^{-} $

Alternatively, out of the three possible oxidation reactions occurring at the anode, i.e.,

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+e^{-},$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}+\frac{1}{2} \mathrm{O}_2+e^{-} $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}_3+e^{-} $

$Ag $ has highest oxidation potential. Hence, $Ag$ of anode is oxidized to $\mathrm{Ag}^{+}$ ions which pass into the solution.

(ii) Electrolysis of aqueous solution of $\mathrm{AgNO}_3$ with platinium electrodes.

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{AgNO}_3(s)+a q \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_3^{-}(a q) $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} $

At Cathode : $\mathrm{Ag}^{+}$ ions have lower discharge potential than $\mathrm{H}^{+}$ ions. Hence, $\mathrm{Ag}^{+}$ ions will be deposited as Ag in preference to $\mathrm{H}^{+}$ ions.

Alternatively, we have standard reduction potentials as :

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s), \mathrm{E}^{\circ}=+0.80 \mathrm{~V}$

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{H}^{+}(a q)+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(g), \mathrm{E}^{\circ}=0.00 \mathrm{~V} $

As $\mathrm{Ag}^{+}$ ions have higher standard reduction potential than that of $\mathrm{H}^{+}$ ions, hence $\mathrm{Ag}^{+}$ ions will be reduced more easily and deposited as Ag.

At Anode : As anode is not attackable, out of $\mathrm{OH}^{-}$ and $\mathrm{NO}_3^{-}$ ions, $\mathrm{OH}^{-}$ ions have lower discharge potential. Hence, $\mathrm{OH}^{-}$ ions will be discharged in preference to $\mathrm{NO}_3^{-}$ ions, which then decompose to give out $\mathrm{O}_2$.

$ \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{OH}+e^{-}, \quad 4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g) $

(iii) Electrolysis of dilute $\mathrm{H}_2 \mathrm{SO}_4$ with platinum electrodes.

$ \begin{aligned} & \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-}(a q) \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \end{aligned} $

At Cathode : $\mathrm{H}^{+}+e \longrightarrow \mathrm{H}, \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2(\mathrm{~g})$

At Anode : $\quad \mathrm{OH}^{-} \longrightarrow \mathrm{OH}+e^-,$

$ \quad\quad\quad\quad\quad 4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2(g)$

Thus, $\mathrm{H}_2$ is liberated at the cathode and $\mathrm{O}_2$ at the anode.

(iv) Electrolysis of aqueous solution of $\mathrm{CuCl}_2$ with platinum electrodes

$ \begin{aligned} & \mathrm{CuCl}_2(s)+a q \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \end{aligned} $

At Cathode : $\mathrm{Cu}^{2+}$ ions will be reduced in preference to $\mathrm{H}^{+}$ ions

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

At Anode : $\mathrm{Cl}^{-}$ ions will be oxidized in preference to $\mathrm{OH}^{-}$ ions

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}+e^{-}, $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_2(g) $

Thus, Cu will be deposited on the cathode and $\mathrm{Cl}_2$ will be liberated at the anode.