Electrochemistry

Electrolysis of NaCl produces NaOH at the cathode.

Charge passed $=4 \times 1200=4800 \mathrm{C}$

Moles of electrons $=\frac{4800}{96500} \approx 0.04974 \mathrm{~mol}$

Since 2 moles of electron produce 2 moles NaOH , NaOH formed $\boldsymbol{\approx} 0.04974 \mathrm{~mol}$

$ \begin{gathered} \text { Concentration }=\frac{0.04974}{0.5}=0.0995 \mathrm{M} \\ \text { pOH }=-\log 0.0995 \\ \text { pOH } \approx 1, \text { so } \mathrm{pH} \approx 14-1=13 \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \mathrm{Ca}^{2+}+2 e^{-} \quad \longrightarrow \mathrm{Ca} \text { (at cathode) } \\ & 1 \text { mole of } \mathrm{CaCl}_2=2 \mathrm{~F} \\ & \,\,\,\,\,\,\, \quad Q=I t \Rightarrow n \cdot 2 F=I t \\ & \Rightarrow \quad t=\frac{n \cdot 2 F}{I}=\frac{20}{40} \times \frac{2 \times 96500}{4} \\ &  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=24125 \mathrm{~s}=6.7 \text { hours. } \end{aligned} \end{aligned} $

The pH of a 0.5 L, 1.0 M NaCl solution after electrolysis for 965 seconds using a 5.0 A current can be determined through the following steps:

Reaction Involved:

$ \text{NaCl} + \text{H}_2\text{O} \xrightarrow{\text{Electrolysis}} \frac{1}{2} \text{H}_2 + \text{Cl}_2 + \text{NaOH} $

Amount of NaCl Present:

In a 0.5 L solution of 1.0 M NaCl, the number of moles of NaCl is $0.5 \text{ mol}$.

Quantity of Electricity Passed:

Current = 5.0 A

Time = 965 s

Total charge passed = $965 \times 5 \text{ coulombs} = 4825 \text{ coulombs}$.

Decomposition of NaCl:

Using Faraday's laws, the moles of NaCl decomposed are calculated as:

$ \frac{4825 \text{ C}}{96500 \text{ C/mol}} = 0.05 \text{ mol} $

Formation of NaOH:

The number of moles of NaOH formed will also be $0.05 \text{ mol}$.

Volume of Solution:

Volume = 0.5 L

Molarity of NaOH:

$ \text{Molarity of NaOH} = \frac{0.05 \text{ moles}}{0.5 \text{ L}} = 0.1 \text{ M} = 10^{-1} \text{ M} $

pOH Calculation:

$ \text{pOH} = 1 $

pH Calculation:

$ \text{pH} = 14 - \text{pOH} = 14 - 1 = 13 $

Thus, after electrolysis, the pH of the solution will be 13.

To find the molarity of the NaOH solution, let's go through the calculations step by step:

Understanding the Fuel Cell Reaction:

The overall reaction for the hydrogen-oxygen fuel cell is:

$ 2 \mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} $

From this reaction, we see that 4 Faradays (F) of charge will produce 2 moles of water. One Faraday is approximately 96,500 Coulombs (C), so 4 F equals 4 × 96,500 C.

Calculating Charge Passed:

The charge ($ Q $) is given by the current ($ I $) multiplied by the time ($ t $). Here, the current is 1 A, and the time is 595.1 hours. Convert this time into seconds:

$ t = 595.1 \times 60 \times 60 \, \text{seconds} $

$ Q = 1 \times 595.1 \times 60 \times 60 = 2,142,360 \, \text{C} $

Determine Water Produced:

Use the relationship between charge available and water produced:

$ \text{Water produced} = \frac{36 \, \text{g}}{4 \times 96,500 \, \text{C}} \times 2,142,360 \, \text{C} \approx 200 \, \text{g} $

This is equivalent to 200 mL of water because the density of water is 1 g/mL.

Calculate Molarity of NaOH Solution:

The NaOH is dissolved in this water. We know:

Mass of NaOH = 5 g

Molar mass of NaOH = 40 g/mol

Volume of water in liters = 200 mL = 0.200 L

Calculate moles of NaOH:

$ \text{Moles of NaOH} = \frac{5 \, \text{g}}{40 \, \text{g/mol}} = 0.125 \, \text{mol} $

Finally, calculate molarity:

$ \text{Molarity} = \frac{\text{moles of NaOH}}{\text{volume in liters}} = \frac{0.125 \, \text{mol}}{0.200 \, \text{L}} = 0.625 \, \text{M} $

Thus, the molarity of the NaOH solution is 0.625 M.

To find the equilibrium constant $K_{eq}$ for the given reaction, we can use the relation between the standard cell potential $E_{\text{cell}}^{\circ}$ and the equilibrium constant. The Nernst equation at standard conditions can be written as:

$E_{\text{cell}}^{\circ} = \frac{0.059}{n} \log K_{eq}$

Here:

   $E_{\text{cell}}^{\circ} = 0.0295\, \text{V}$

   $n = 2$ (number of electrons transferred in the balanced reaction)

   $\frac{2.303RT}{F} = 0.059$

Now, solving for $K_{eq}$:

$0.0295 = \frac{0.059}{2} \log K_{eq}$

Multiplying both sides by 2 to isolate the logarithm term:

$0.059 = 0.059 \log K_{eq}$

Dividing both sides by 0.059:

$1 = \log K_{eq}$

Taking the antilog (base 10) of both sides:

$K_{eq} = 10^1 = 10$

So, the correct answer is Option D: 10.

(a) 2Fe³⁺ + 2e^$-$ $\to$ 2Fe²⁺ ; E$^\circ$ = 0.77 V

2I^$-$ $\to$ I₂ + 2e^$-$ ; E$^\circ$ = $-$0.54 V (sign of E$^\circ$ is reversed)

2Fe³⁺ + 2I^$-$ $\to$ 2Fe²⁺ + I₂ ; E$_{cell}^o$ = + 0.23 V

This reaction is feasible, since E$_{cell}^o$ is positive.

(b) Cu $\to$ Cu²⁺ + 2e^$-$ ; E$^\circ$ = $-$ 0.34 V (sign of E$^\circ$ has been reversed)

2Ag⁺ + 2e^$-$ $\to$ 2Ag ; E$^\circ$ = + 0.80 V

Cu + 2Ag⁺ $\to$ Cu²⁺ + 2Ag ; E$^\circ$ = + 0.46 V

This reaction is feasible, since E$_{cell}^o$ is positive.

(c) 2Fe³⁺ + 2e^$-$ $\to$ 2Fe²⁺ ; E$^\circ$ = + 0.77 V

Cu $\to$ Cu²⁺ + 2e^$-$ ; E$^\circ$ = $-$ 0.34 V (sign of E$^\circ$ is reversed)

2Fe³⁺ + Cu $\to$ 2Fe²⁺ + Cu²⁺ ; E$^\circ$ = + 0.43 V

This reaction is feasible, since E$_{cell}^o$ is positive.

(d) Ag $\to$ Ag⁺ + e^$-$ ; E$^\circ$ = $-$ 0.80 V (sign of E$^\circ$ is reversed)

Fe³⁺ + e^$-$ $\to$ Fe²⁺ ; E$^\circ$ = + 0.77 V

Ag + Fe³⁺ $\to$ Ag⁺ + Fe²⁺ ; E$^\circ$ = $-$ 0.03 V

This reaction is not feasible, since E$_{cell}^o$ is negative.

NaCl + KNO₃ $\to$ NaNO₃ + KCl

Sum of molar conductivity of reactant = Sum of molar conductivity of products

$\Lambda ^\circ (NaN{O_3}) = \Lambda _m^o(NaCl) + \Lambda _m^o(KN{O_3}) - \Lambda _m^o(KCl)$

$KCl - NaCl = - \Lambda _m^oKCl - \Lambda _m^o(NaCl) = 24$

Molar conductivities of KNO₃ = 111

Molar conductivities of NaNO₃ = 111 $-$ 24 = 87 S cm² mol^$-$1

Ca²⁺ + 2e^$-$ $\to$ Ca (at cathode)

1 mole CaCl₂ = 2F

From Q = it $\Rightarrow$ n . 2F = it

$ \Rightarrow t = {{n\,.\,2F} \over i} = {{30} \over {40}} \times {{2 \times 96500} \over 5}$

= 28950 s $ = {{28960} \over {60 \times 60}} = 8.04$ = 8 h

As the value of reduction potential of metal ion increases, the tendency of metal oxide to get reduce into metal increases.

Since, reduction potential of only Ag is positive among the given, thus Ag₂O readily gets reduced to Ag metal. In other words, it can be said that Ag₂O is the least stable oxide among the given.

According to Faraday's first law of electrolysis:

The reaction at cathode : $\mathop {C{u^{2 + }}}\limits_{63.5\,g} + 2{e^ - } \to \mathop {Cu}\limits_{2 \times 96500C} $

The quantity of charge passed

= I $\times$ t = (10 amp) $\times$ (60 $\times$ 60s) = 36000 C.

2 $\times$ 96500 C of charge deposit copper - 63.5 g

36000 C of charge deposit copper

$ = {{(63.5\,g)} \over {(2 \times 96500\,C)}} \times (36000C)$ = 11.84 g

Thus, 11.84 g of copper will dissolve from the anode and the same amount from the solution will get deposited on the cathode. The concentration of the solution will remain unchanged.