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

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of $\mathrm{K}^{+} / \mathrm{K}^{2} \mathrm{Mg}^{2+} / \mathrm{Mg}<\mathrm{Cr}^{3+} / \mathrm{Cr}<\mathrm{Hg}^{2+} / \mathrm{Hg}<\mathrm{Ag}^{+} / \mathrm{Ag}$.

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

$\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K}$

Answer

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

$\mathrm{Zn_{(s)}} |\mathrm{Zn_{(a q)}^{2+}} \|\mathrm{Ag_{(a q)}^{+}} | \mathrm{Ag_{(s)}}$

(i) Zn electrode (anode) is negatively charged.

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

(iii) The reaction taking place at the anode is given by,

$\mathrm{Zn_{(s)}} \longrightarrow \mathrm{Zn_{(a q)}^{2+}}+2 \mathrm{e}^{-}$

The reaction taking place at the cathode is given by,

$\mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag_{(s)}}$

Answer

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

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

The galvanic cell of the given reaction is depicted as:

$\mathrm{Cr_{(s)}} |\mathrm{Cr_{(a q)}^{3+}}||\mathrm{Cd_{(a q)}^{2+}}| \mathrm{Cd_{(s)}}$

Now, the standard cell potential is

$$ \begin{aligned} E_{\text {cell }}^{\ominus} & =E_{\mathrm{R}}^{\ominus}-E_{\mathrm{L}}^{\ominus} \\ & =-0.40-(-0.74) \\ & =+0.34 \mathrm{~V} \\ \Delta_{\mathrm{r}} G^{\ominus} & =-n \mathrm{~F} E_{\text {cell }}^{\ominus} \end{aligned} $$

In the given equation, $n=6$

$\mathrm{F}=96487 \mathrm{C} \mathrm{mol}^{-1}$

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

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

$=-196833.48 \mathrm{CV} \mathrm{mol}^{-1}$

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

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

Again,

$\Delta_{\mathrm{r}} G^{\ominus}=-\mathrm{R} T \ln K$

$\Rightarrow \Delta_{\mathrm{r}} G^{\ominus}=-2.303 \mathrm{R} T \ln K$

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

$$ =\frac{-196.83 \times 10^{3}}{2.303 \times 8.314 \times 298} $$

$=34.496$

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

$=3.13 \times 10^{34}$

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

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

The galvanic cell of the given reaction is depicted as:

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

Now, the standard cell potential is

$$ \begin{aligned} E_{\text {cell }}^{\ominus} & =E_{\mathrm{R}}^{\ominus}-E_{\mathrm{L}}^{\ominus} \\ & =0.80-0.77 \\ & =0.03 \mathrm{~V} \end{aligned} $$

Here, $n=1$.

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

$=-1 \times 96487 \mathrm{C} \mathrm{mol}^{-1} \times 0.03 \mathrm{~V}$

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

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

$$ \begin{aligned} & \text { Again, } \Delta_{\mathrm{r}} G^{\ominus}=-2.303 \mathrm{R} T \ln K \\ & \Rightarrow \log K=-\frac{\Delta_{\mathrm{r}} G}{2.303 \mathrm{R} T} \\ & \quad=\frac{-2894.61}{2.303 \times 8.314 \times 298} \\ & =0.5073 \\ & \therefore \mathrm{K}=\text { antilog (0.5073) } \\ & =3.2 \text { (approximately) } \end{aligned} $$

Answer

(i) For the given reaction, the Nernst equation can be given as:

$$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} \\ & ={0.34-(-2.36)}-\frac{0.0591}{2} \log \frac{.001}{.0001} \\ & =2.7-\frac{0.0591}{2} \log 10 \end{aligned} $$

$=2.7-0.02955$

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

(ii) For the given reaction, the Nernst equation can be given as:

$$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \\ & \quad={0-(-0.44)}-\frac{0.0591}{2} \log \frac{0.001}{1^{2}} \\ & \quad=0.44-0.02955(-3) \\ & =0.52865 \mathrm{~V} \end{aligned} $$

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

(iii) For the given reaction, the Nernst equation can be given as:

$$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \\ & ={0-(-0.14)}-\frac{0.0591}{2} \log \frac{0.050}{(0.020)^{2}} \end{aligned} $$

$=0.14-0.0295 \times \log 125$

$=0.14-0.062$

$=0.078 \mathrm{~V}$

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

(iv) For the given reaction, the Nernst equation can be given as:

$$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{1}{\left[\mathrm{Br}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{2}} \\ & =(0-1.09)-\frac{0.0591}{2} \log \frac{1}{(0.010)^{2}(0.030)^{2}} \\ & =-1.09-0.02955 \times \log \frac{1}{0.00000009} \\ & =-1.09-0.02955 \times \log \frac{1}{9 \times 10^{-8}} \\ & =-1.09-0.02955 \times \log \left(1.11 \times 10^{7}\right) \\ & =-1.09-0.02955(0.0453+7) \\ & =-1.09-0.208 \\ & =-1.298 \mathrm{~V} \end{aligned} $$

Answer

$$ \begin{array}{rl} \mathrm{Zn_{(s)}}& \longrightarrow \mathrm{Zn_{(a q)}^{2+}}+2 \mathrm{e}^{-} ; E^{\ominus}=0.76 \mathrm{~V} \\ \mathrm{Ag_2} \mathrm{O_{(s)}}+\mathrm{H_2} \mathrm{O_{(b)}}+2 \mathrm{e}^{-}&\longrightarrow 2 \mathrm{Ag_{(s)}}+2 \mathrm{OH_{(a q)}^{-}} ; E^{\ominus}=0.344 \mathrm{~V}\\ \hline \mathrm{Zn_{(s)}}+\mathrm{Ag_2} \mathrm{O_{(s)}}+\mathrm{H_2} \mathrm{O_{(b)}}& \longrightarrow \mathrm{Zn_{(a q)}^{2+}}+2 \mathrm{Ag_{(s)}}+2 \mathrm{OH_{(a q)}^{-}} ; E^{\ominus}=1.104 \mathrm{~V} \end{array} $$

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

We know that,

$ \begin{aligned} & \Delta_{r} G^{\ominus}=-n \mathrm{~F} E^{\ominus} \\ = & -2 \times 96487 \times 1.04 \\ = & -213043.296 \mathrm{~J} \\ = & -213.04 \mathrm{~kJ} \end{aligned} $

Answer

Conductivity of a solution is defined as the conductance of a solution of $1 \mathrm{~cm}$ in length and area of cross-section $1 \mathrm{sq}$. $\mathrm{cm}$. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol $\mathring{A}$. If $\mathring{A}$ is resistivity, then we can write:

$$ \kappa=\frac{1}{\rho} $$

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.

i.e.,

$$ G=\kappa \frac{a}{l}=\kappa \cdot 1=\kappa $$

(Since $a=1, I=1$ )

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 $\mathrm{V}$ 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.

$$ \Lambda_{m}=\kappa \frac{A}{l} $$

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

$$ \therefore \Lambda_{m}=\kappa \mathrm{V} $$

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 $\Lambda_{m}$ with $\sqrt{c}$ for strong and weak electrolytes is shown in the following plot:

Answer

Given,

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

$\mathrm{C}=0.20 \mathrm{M}$

$\therefore$ Molar conductivity,

$$ \Lambda_{m}=\frac{\kappa \times 1000}{\mathrm{c}} $$

$=\frac{0.0248 \times 1000}{0.2}$

$=124 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

Answer

Given,

Conductivity, $\mathring{A}=0.146 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$

Resistance, $R=1500 \mathring{A}$

$\therefore$ Cell constant $=\mathring{A} \times R$

$=0.146 \times 10^{-3} \times 1500$

$=0.219 \mathrm{~cm}^{-1}$

Answer

Given,

$\mathring{A}=1.237 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}, \mathrm{c}=0.001 \mathrm{M}$

Then, $\mathring{A}=1.237 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}, \mathrm{c} 1 / 2=0.0316 \mathrm{M}^{1 / 2}$

$\therefore \Lambda_{m}=\frac{\kappa}{c}$

$=\frac{1.237 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.001 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

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

Given,

$\mathring{A}=11.85 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}, \mathrm{c}=0.010 \mathrm{M}$

Then, $\mathring{A}=11.85 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}, \mathrm{c}^{1 / 2}=0.1 \mathrm{M}^{1 / 2}$

$\therefore \Lambda_{m}=\frac{\kappa}{c}$

$=\frac{11.85 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.010 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

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

Given,

$\mathring{A}=23.15 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}, \mathrm{c}=0.020 \mathrm{M}$

Then, $\mathring{A}=23.15 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}, \mathrm{c}^{1 / 2}=0.1414 \mathrm{M}^{1 / 2}$

$\therefore \Lambda_{m}=\frac{\kappa}{c}$ $=\frac{23.15 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.020 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

$=115.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Given,

$\mathring{A}=55.53 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}, \mathrm{c}=0.050 \mathrm{M}$

Then, $\mathring{A} = 55.53 \times 10^{-4} \mathrm{S cm^{-1}}, \mathrm{c^{\frac{1}{2}}} = 0.2236 \mathrm{ M^{\frac{1}{2}}}$

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

$=\frac{55.53 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.050 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

$=111.11 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Given,

$\mathring{A}=106.74 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}, \mathrm{c}=0.100 \mathrm{M}$

Then, $\mathring{A}=106.74 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}, \mathrm{c}^{1 / 2}=0.3162 \mathrm{M}^{1 / 2}$

$\therefore \Lambda_{m}=\frac{\kappa}{c}$

$=\frac{106.74 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.100 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$

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

Now, we have the following data:

Answer

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

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

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

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

$=32.76 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

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

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

$=0.084$

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

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

$=1.86 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$

Answer

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

$\therefore$ Required charge $=3 \mathrm{~F}$

$=3 \times 96487 \mathrm{C}$

$=289461 \mathrm{C}$

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

$\therefore$ Required charge $=2 \mathrm{~F}$

$=2 \times 96487 \mathrm{C}$

$=192974 \mathrm{C}$

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

i.e., $\mathrm{Mn}^{7+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}$ $\therefore$ Required charge $=5 \mathrm{~F}$

$=5 \times 96487 \mathrm{C}$

$=482435 \mathrm{C}$

Answer

(i) According to the question,

$$ \mathrm{Ca}^{2+}+2 \mathrm{e}^{-1} \longrightarrow \underset{40 \mathrm{~g}}{\mathrm{Ca}} $$

Electricity required to produce $40 \mathrm{~g}$ of calcium $=2 \mathrm{~F}$

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

$=1 \mathrm{~F}$

(ii) According to the question,

$$ \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \underset{27 \mathrm{~g}}{\mathrm{Al}} $$

Electricity required to produce $27 \mathrm{~g}$ of $\mathrm{Al}=3 \mathrm{~F}$

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

$=4.44 \mathrm{~F}$

Answer

(i) 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 $1 \mathrm{~mol}$ of $\mathrm{H_2} \mathrm{O}$ to $\mathrm{O_2}=2 \mathrm{~F}$

$=2 \times 96487 \mathrm{C}$

$=192974 \mathrm{C}$

(ii) According to the question,

$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1}$

Electricity required for the oxidation of $1 \mathrm{~mol}$ of $\mathrm{FeO}$ to $\mathrm{Fe_2} \mathrm{O_3}=1 \mathrm{~F}$

$=96487 \mathrm{C}$

Answer

Given,

Current $=5 \mathrm{~A}$

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

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

$=5 \times 1200$

$=6000 \mathrm{C}$

According to the reaction,

$\mathrm{Ni_{(a q)}^{2+}}+2 \mathrm{e}^{-} \longrightarrow \underset{58.7 \mathrm{~g}}{\mathrm{Ni_{(s)}}}$

Nickel deposited by $2 \times 96487 \mathrm{C}=58.71 \mathrm{~g}$

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

$=1.825 \mathrm{~g}$

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

Answer

According to the reaction:

$\mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \underset{108 \mathrm{~g}}{\mathrm{Ag_{(s)}}}$

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 $=\frac{96487 \times 1.45}{108} \mathrm{C}$

$=1295.43 \mathrm{C}$

Given,

Current $=1.5 \mathrm{~A}$

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

$=863.6 \mathrm{~s}$

$=864 \mathrm{~s}$

$=14.40 \mathrm{~min}$

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

$$ =\frac{63.5 \times 1295.43}{2 \times 96487} \mathrm{~g} $$

$=0.426 \mathrm{~g}$ of $\mathrm{Cu}$

$$ \begin{array}{r} \mathrm{Zn_{(a q)}^{2+}}+2 \mathrm{e}^{-} \longrightarrow \underset{65.4 \mathrm{~g}}{\mathrm{Zn_{(s)}}} \\ \end{array} $$

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 $=\frac{65.4 \times 1295.43}{2 \times 96487} \mathrm{~g}$

$=0.439 \mathrm{~g}$ of $\mathrm{Zn}$

Answer

(i)

$$ \begin{array}{clr} [\mathrm{Fe _{(\mathrm{aq})}^{3+}}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe _{(\mathrm{aq})}^{2+}}] \times 2 ; & E^{0}=+0.77 \mathrm{~V} \\ 2 \mathrm{I _{(\mathrm{aq})}^{-}} \longrightarrow \mathrm{I _{2(\mathrm{~s})}}+2 \mathrm{e}^{-} ; & E^{0}=-0.54 \mathrm{~V} \\ \hline 2 \mathrm{Fe _{(\mathrm{aq})}^{3+}}+2 \mathrm{I _{(\mathrm{aq})}^{-}} \longrightarrow 2 \mathrm{Fe _{(\mathrm{aq})}^{2+}}+\mathrm{I _{2(\mathrm{~s})}} ; & E^{0}=+0.23 \mathrm{~V} \end{array} $$

Since $E^{\circ}$ for the overall reaction is positive, the reaction between $\mathrm{Fe_{(a q)}^{3+}} $ and $\mathrm{I_{(a)}^{-}}$ is feasible.

(ii)

$$ \begin{array}{lcl} \mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-}& \longrightarrow \left. \mathrm{Ag_{(s)}}\right] \times 2 ; & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Cu_{(s)}}& \longrightarrow \mathrm{Cu_{(a q)}^{2+}}+2 \mathrm{e}^{-} ; & E^{\circ}=-0.34 \mathrm{~V} \\ \hline 2 \mathrm{Ag_{(a q)}^{+}}+\mathrm{Cu_{(s)}}& \longrightarrow 2 \mathrm{Ag_{(s)}}+\mathrm{Cu_{(a q)}^{2+}} ; & E^{\circ}=+0.46 \mathrm{~V} \end{array} $$

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

(iii) $$ \begin{array}{lcl} \mathrm{Fe_{(a q)}^{3+}}+\mathrm{e}^{-} & \longrightarrow \left. \mathrm{Fe_{(a q)}^{2+}}\right] \times 2 ; & E^{0}=+0.77 \mathrm{~V}\\ 2 \mathrm{Br_{(aq)}^{-}} & \longrightarrow \mathrm{Br_{2(l)}}+2 \mathrm{e}^{-} ; & E^{0}=-1.09 \mathrm{~V} \\ \hline 2 \mathrm{Fe_{(\mathrm{aq})}^{3+}}+2 \mathrm{Br_{(\mathrm{aq})}^{-}} & \longrightarrow 2 \mathrm{Fe_{(\mathrm{aq})}^{2+}} \text { and } \mathrm{Br_{2(l)}} ; & E^{\circ}=-0.32 \mathrm{~V} \end{array} $$

Since $E^{0}$ for the overall reaction is negative, the reaction between $\mathrm{Fe_{(a q)}^{3+}}$ and $\mathrm{Br_{(a q)}^{-}}$ is not feasible.

(iv)

$$ \begin{array}{lll} \mathrm{Ag_{(s)}} &\longrightarrow \mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-} \quad ; & E^{0}=-0.80 \mathrm{~V} \\ \mathrm{Fe_{(a q)}^{3+}}+\mathrm{e}^{-}& \longrightarrow \mathrm{Fe_{(a q)}^{2+}} \quad ; & E^{0}=+0.77 \mathrm{~V} \\ \hline \mathrm{Ag_{(s)}}+\mathrm{Fe_{(a q)}^{3+}} & \longrightarrow \mathrm{Ag_{(a q)}^{+}}+\mathrm{Fe_{(a q)}^{2+}} \quad ; & E^{0}=-0.03 \mathrm{~V} \end{array} $$

Since $E^{\mathrm{o}} \mathrm{E}$ for the overall reaction is negative, the reaction between $\mathrm{Ag_{(s)}}$ and $\mathrm{Fe_{(a q)}^{3+}}$ is not feasible.

(iv) $$ \begin{array}{lll} \mathrm{Br_{2(a q)}}+2 \mathrm{e}^{-} &\longrightarrow 2 \mathrm{Br_{(aq)}^{-}} \quad ; & E^{0}=+1.09 \mathrm{~V}\\ \mathrm{Fe_{(\mathrm{aq})}^{2+}} &\longrightarrow \left. \mathrm{Fe_{(\mathrm{aq})}^{3+}}+\mathrm{e}^{-}\right] \times 2 ; &E^{0}=-0.77 \mathrm{~V} \\ \hline \mathrm{Br_{2(\mathrm{aq})}}+2 \mathrm{Fe_{(\mathrm{aq})}^{2+}} &\longrightarrow 2 \mathrm{Br_{(\mathrm{aq})}^{-}}+2 \mathrm{Fe_{(\mathrm{aq})}^{3+}} \quad ; & E^{0}=+0.32 \mathrm{~V} \end{array} $$

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) At cathode:

The following reduction reactions compete to take place at the cathode.

$$ \begin{aligned} & \mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag_{(s)}} ; E^{0}=0.80 \mathrm{~V} \\ & \mathrm{H_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H_{2(g)}} ; E^{0}=0.00 \mathrm{~V} \end{aligned} $$

The reaction with a higher value of $E^{0}$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The $\mathrm{Ag}$ anode is attacked by $\mathrm{NO_3}^{-}$ions. Therefore, the silver electrode at the anode dissolves in the solution to form $\mathrm{Ag}^{+}$.

(ii) At cathode:

The following reduction reactions compete to take place at the cathode.

$$ \begin{aligned} & \mathrm{Ag_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag_{(s)}} ; E^{0}=0.80 \mathrm{~V} \\ & \mathrm{H_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H_{2(g)}} ; E^{\circ}=0.00 \mathrm{~V} \end{aligned} $$

The reaction with a higher value of $E^{\circ}$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

Since Pt electrodes are inert, the anode is not attacked by $\mathrm{NO_3}^{-}$ions. Therefore, $\mathrm{OH}^{-}$or $\mathrm{NO_3}^{-}$ions can be oxidized at the anode. But $\mathrm{OH}$ ’ ions having a lower discharge potential and get preference and decompose to liberate $\mathrm{O_2}$.

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

$4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H_2} \mathrm{O}+\mathrm{O_2}$

(iii) At the cathode, the following reduction reaction occurs to produce $\mathrm{H_2}$ gas.

$\mathrm{H_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H_{2(g)}}$

At the anode, the following processes are possible.

$$ \begin{align*} & 2 \mathrm{H_2} \mathrm{O_{(l)}} \longrightarrow \mathrm{O_{2(g)}}+4 \mathrm{H_{(a q)}^{+}}+4 \mathrm{e}^{-} ; E^{\circ}=+1.23 \mathrm{~V} \tag{i}\\ & 2 \mathrm{SO_{4(a q)}^{2-}} \longrightarrow \mathrm{S_2} \mathrm{O_{6(a q)}^{2-}}+2 \mathrm{e}^{-} ; E^{\circ}=+1.96 \mathrm{~V} \tag{ii} \end{align*} $$

For dilute sulphuric acid, reaction (i) is preferred to produce $\mathrm{O_2}$ gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At cathode:

The following reduction reactions compete to take place at the cathode.

$$ \begin{aligned} & \mathrm{Cu_{(a q)}^{2+}}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu_{(s)}} ; E^{0}=0.34 \mathrm{~V} \\ & \mathrm{H_{(a q)}^{+}}+\mathrm{e}^{-} \longrightarrow 1 / 2 \mathrm{H_{2(g)}} ; E^{0}=0.00 \mathrm{~V} \end{aligned} $$

The reaction with a higher value of $E^{0}$ takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

At anode:

The following oxidation reactions are possible at the anode.

$$ \begin{aligned} & \mathrm{Cl_{(a q)}^{-}} \longrightarrow 1 / 2 \mathrm{Cl_{2(g)}}+\mathrm{e}^{-1} ; E^{\circ}=1.36 \mathrm{~V} \\ & 2 \mathrm{H_2} \mathrm{O_{(l)}} \longrightarrow \mathrm{O_{2(g)}}+4 \mathrm{H_{(a q)}^{+}}+4 \mathrm{e}^{-} ; E^{\circ}=+1.23 \mathrm{~V} \end{aligned} $$

At the anode, the reaction with a lower value of $E^{0}$ is preferred. But due to the over-potential of oxygen, $\mathrm{Cl}^{-}$ gets oxidized at the anode to produce $\mathrm{Cl_2}$ gas.