Electrochemistry

To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine ($\mathrm{N}_2 \mathrm{H}_4$) by $\mathrm{O}_2$, let's analyze each option in detail.

Option A: $\mathrm{OH}^{-}$ ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2$$(\mathrm{~g})$ and water, releasing 4 electrons to the anode. This statement is accurate because in an electrochemical cell, the anode is where oxidation occurs. Hydrazine can be oxidized in the presence of $\mathrm{OH}^{-}$ ions to yield nitrogen gas and water, releasing electrons as a part of this redox reaction:

Option B: At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2$ $(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water. This statement is incorrect because, usually, the cathode is where reduction occurs, not a breaking down of hydrazine. Instead, reduction reactions involve the gain of electrons. Additionally, at the cathode, it's more common for oxygen to be reduced rather than hydrazine itself breaking down.

Option C: At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$. This statement is correct. At the cathode in an alkaline medium, oxygen undergoes reduction to form hydroxide ions $\mathrm{OH}^{-}$, as depicted by the half-reaction:

Option D: Oxides of nitrogen are major by-products of the electrochemical process. This statement is incorrect because the primary product from the electrochemical reaction of hydrazine is nitrogen gas ($\mathrm{N}_2$), and not oxides of nitrogen. The formation of oxides of nitrogen would be notable only under different specific conditions not mentioned in this context.

Therefore, the correct statements regarding the electrochemical oxidation of hydrazine by $\mathrm{O}_2$ are:

Option A and Option C.

Option (P) :

On adding $\mathrm{AgNO}_3$ solution to $\mathrm{KCl}$ solution precipitation of $\mathrm{AgCl}$ will occur due to which $\mathrm{Cl}^{-}$already present will be replaced by $\mathrm{NO}_3^{-}$ions. So conductance of solution will decrease till equivalence point. After complete precipitation of $\mathrm{AgCl}$, further added $\mathrm{AgNO}_3$ will increase the number of ions in resulting solution so conductance will increase.

Option (Q) :

On adding $\mathrm{KCl}$ solution to $\mathrm{AgNO}_3$ solution precipitation of $\mathrm{AgCl}$ will occur due to which already present $\mathrm{Ag}^{+}$ions will be replaced by $\mathrm{K}^{+}$ions in solution. So conductance of solution will increase. After complete precipitation of $\mathrm{AgCl}$ further added $\mathrm{KCl}$ will increase the number of ions in resulting solution so conductance will increase further.

Option (R) :

On adding $\mathrm{HCl}$ solution to $\mathrm{NaOH}$ solution, $\mathrm{OH}^{-}$will be replaced by $\mathrm{Cl}^{-}$ions so conductance of solution decreases. After complete neutralisation further added $\mathrm{HCl}$ will increase number of ions in the solution. So conductance will increase futher.

Option (S) :

On adding $\mathrm{CH}_3 \mathrm{COOH}$ solution to $\mathrm{NaOH}$ solution $\mathrm{OH}^{-}$will be replaced by $\mathrm{CH}_3 \mathrm{COO}^{-}$ions, so conductance of solution decreases. After complete neutralisation further added $\mathrm{CH}_3 \mathrm{COOH}$ will remain undissociated because it is a weak acid and there is also common ion effect on acetate ions. So number of ions in solution will remain almost constant therefore conductance of solution will remain constant.

Given, Mass of copper $ =0.592 \mathrm{~g} $, time $ =60 $ minutes, current $ =0.5 \mathrm{~A} $ Total charge passes through solution,

$ \begin{array}{l} Q=I \times t \\\\ Q=0.5 \times 60 \times 60=1800 \mathrm{C} \end{array} $

Number of moles of electrons transferred.

$ \begin{array}{l} n=\frac{Q}{F}=\frac{1800 \mathrm{C}}{96500 \mathrm{C} / \mathrm{mol}} \\\\ \Rightarrow \quad n=0.01865 \mathrm{~mol} \end{array} $

$ \mathrm{Cu}^{2+} $ requires 2 moles of electron to deposit 1 mole of copper

$ \begin{array}{c} \text { Moles of copper }=\frac{n}{2}=\frac{0.01865}{2} \\\\ =0.009325 \mathrm{~mol} \end{array} $

Mass of copper deposited

$ \begin{array}{l} =0.009325 \times 63.55 \\\\ =0.592 \mathrm{~g} \end{array} $

Electrochemical equivalent is given by

$ \frac{\text { Mass of copper deposited }}{\text { Charged passed }}=\frac{0.592}{180} $

$ =3.3 \times 10^{-4} \mathrm{~g} / \mathrm{C} $

The standard electrode potentials for the $\mathrm{Li}^{+} / \mathrm{Li}$ and $\mathrm{Na}^{+} / \mathrm{Na}$ half-cells are as follows:

$\mathrm{Li}^{+} / \mathrm{Li}$: $E^{\circ} = -3.04 \, \mathrm{V}$

$\mathrm{Na}^{+} / \mathrm{Na}$: $E^{\circ} = -2.714 \, \mathrm{V}$

Therefore, the correct option is:

Option A

$ -3.04, -2.714 $

These values indicate the tendency of lithium and sodium to lose electrons and form cations, with lithium being more reactive in this context due to its more negative electrode potential.

Statement $ I $ is correct but statement II is incorrect. Correct form of Statement II is

At cathode (Reduction take place)

$ 2 \mathrm{Cu}^{2+}+4 e^{-} \xrightarrow{}2 \mathrm{Cu} $

At anode (oxidation)

$ 4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-} $

Hence, oxygen is liberated at anode not cathode.

Given,

Molar conductivity

$ =124 \times 10^{-4} \mathrm{Sm}^{2} / \mathrm{mol} $

Molarity $ =0.02 \mathrm{M} $,

Cell constant $ =129 \mathrm{~m}^{-1} $

Molar conductivity is given by

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

where $ \kappa $ is conductivity

$ \mathrm{K}=\frac{\Lambda_{\mathrm{m}} \times M}{1000} $

Substitute the vaues in Eq. (i),

$ \kappa=\frac{124 \times 10^{-4} \times 0.02}{1000} \Rightarrow \kappa=0.248 \ldots $

Conductivity, k is given by

$ \begin{array}{c} \mathrm{K}=\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \text { or Resistance }=\frac{\text { Cell constant }}{\mathrm{K}} \\\\ R=\frac{129}{0.248} \text { or } R=520 \Omega \end{array} $

Using Kohlrausch's law we get, (for formic acid)

$ \begin{aligned} & \left.\lambda_{(\mathrm{HCOOH})}^{\circ}=\lambda_{\left(\mathrm{H}^{+}\right)}^{\circ}+\lambda_{(\mathrm{HCOO})}^{\circ}\right) \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=349.6+54.6 \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=404.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Degree of dissociation is given by

$ \alpha=\frac{\lambda_{(\mathrm{HCOOH})}}{\lambda_{(\mathrm{HCOOH})}^{\circ}} $

[where, $\alpha=0.11$ (as $11 \%$ given)]

$ \begin{aligned} \lambda_{(\mathrm{HCOOH})} & =\text { Molar conductivity of } \mathrm{HCOOH} \\\\ \lambda_{(\mathrm{HCOOH})} & =\alpha \times \lambda_{(\mathrm{HCOOH})}^{\circ} \\\\ & =0.11 \times 404: 2 \\\\ & =44.46 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Correct Statements

To evaluate the given statements, we analyze each one:

Statement (A): The potential of the hydrogen electrode in a solution of $\text{pH} = 10$ at 298 K.

The potential of a hydrogen electrode can be calculated using the Nernst equation for the hydrogen electrode:

$ E = E^\ominus - \frac{0.0591}{n} \log [\text{H}^+] $

At standard conditions $E^\ominus = 0$, and $[\text{H}^+]$ is calculated from the pH:

$ \text{pH} = 10 \implies [\text{H}^+] = 10^{-10} $

Substitute into the Nernst equation:

$ E = -0.0591 \times \log(10^{-10}) = -0.59 \, \text{V} $

Statement A is correct.

Statement (B): Limiting molar conductivity of $\text{Ca}^{2+}$ and $\text{Cl}^{-}$ are 119 and 76 S cm$^2$ mol$^{-1}$ respectively.

The limiting molar conductivity of $\text{CaCl}_2$ is the sum of the molar conductivities of $\text{Ca}^{2+}$ and two $\text{Cl}^{-}$ ions:

$ \lambda^\circ_{\mathrm{CaCl}_2} = \lambda^\circ_{\mathrm{Ca}^{2+}} + 2\lambda^\circ_{\mathrm{Cl}^-} $

Substitute the given values:

$ \lambda^\circ_{\mathrm{CaCl}_2} = 119 + 2(76) = 119 + 152 = 271 \, \text{S cm}^2 \text{ mol}^{-1} $

The statement given as 195 S cm$^2$ mol$^{-1}$ is incorrect.

Statement (C): Relationship between $K_C$ and $E_{\text{cell}}^{\ominus}$.

The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation:

$ E_{\text{cell}}^{\ominus} = \frac{RT}{nF} \ln K_C $

Converting natural log to base-10 log:

$ E_{\text{cell}}^{\ominus} = \frac{2.303 RT}{nF} \log K_C $

The statement is correctly stated.

Conclusion

Based on the analysis, the correct statements are A and C.

Option C: A, C only is the correct answer.

Given the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

The standard cell potential is:

$ E^{\ominus} = 1 \, \text{V} $

The standard entropy change is:

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $

The temperature is:

$ T = 300 \, \text{K} $

There is a relationship between the change in Gibbs free energy, entropy, and enthalpy, given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Also, the change in Gibbs free energy is related to the cell potential and the number of moles of electrons transferred ($n$) through the equation:

$ \Delta_r G^{\ominus} = -n F E^{\ominus} $

Where $ F $ is the Faraday constant ($96500 \, \text{C/mol}$).

Substituting into the equations, we find:

$ -n F E^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Rearranging and substituting the known values:

$ -2 \times 96500 \times 1 = \Delta_r H^{\ominus} - 300 \times 100 $

Solving this, we get:

$ \Delta_r H^{\ominus} = -163 \times 10^3 \, \text{J/mol} $

Therefore:

$ \Delta_r H^{\ominus} = -163 \, \text{kJ/mol} $

To find the change in entropy ($\Delta_r S^{\ominus}$) for the given cell reaction at 300 K, we start with the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

Given parameters are:

Standard cell potential, $E^{\ominus} = 1.0 \, \text{V}$

Standard reaction enthalpy change, $\Delta_r H^{\ominus} = -163 \, \text{kJ/mol}$

Temperature, $T = 300 \, \text{K}$

Faraday's constant, $F = 96500 \, \text{C/mol}$

The relationship between Gibbs free energy change ($\Delta_r G^{\ominus}$), enthalpy change ($\Delta_r H^{\ominus}$), and entropy change ($\Delta_r S^{\ominus}$) is given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Additionally, the change in Gibbs free energy for the cell is also related to the cell potential by the equation:

$ \Delta_r G^{\ominus} = -nFE^{\ominus} $

where $n$ is the number of moles of electrons exchanged in the reaction. Here, $n = 2$ since two electrons are transferred.

Substituting the known values into the equation:

$ -nFE^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Calculate as follows:

$ -2 \times 96500 \times 1 = -163000 - T \Delta_r S^{\ominus} $

Simplifying:

$ -193000 = -163000 - 300 \Delta_r S^{\ominus} $

Rearranging to solve for $\Delta_r S^{\ominus}$:

$ 30000 = 300 \Delta_r S^{\ominus} $

$ \Delta_r S^{\ominus} = \frac{30000}{300} $

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $

$ \text { The cell reaction is, } $

$ \begin{aligned} & \therefore \quad Q=\frac{\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2}{\mathrm{p}_{\mathrm{Cl}_2}} \\ & \text { and } E_{\text {cell }}=E^{\circ}-\frac{0.0591}{2} \log Q \\ & \text { So, } E_{\text {cell }} \propto \frac{1}{Q} \end{aligned} $

(a) $Q=\frac{1 \times 1^2}{1}=1$;

(b) $Q=\frac{0.01 \times 1^2}{1}=10^{-2}$

(c) $Q=\frac{1 \times(0.01)^2}{1}=10^{-4}$;

(d) $Q=\frac{0.01 \times(0.01)^2}{1}=10^{-6}$

$\Rightarrow$ The order of the values of

$ Q: d

$\Rightarrow$ The order of emf of galvanic cells:

$ d>c>b>a $

The emf is maximum in

$ \begin{aligned} & \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(0.01 \mathrm{M})\right| \\ & \mathrm{Cl}_2(1 \mathrm{~atm}) \mid \mathrm{Pt} \end{aligned} $

Let's analyze the data step by step.

We are given the following standard electrode reactions and their potentials:

For aluminum (Al):

$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}, \quad E^\circ = -1.66 \text{ V}$

$\mathrm{Al}^{+} + e^- \to \mathrm{Al}, \quad E^\circ = +0.55 \text{ V}$

For thallium (Tl):

$\mathrm{Tl}^{3+} + 3e^- \to \mathrm{Tl}, \quad E^\circ = +1.26 \text{ V}$

$\mathrm{Tl}^{+} + e^- \to \mathrm{Tl}, \quad E^\circ = -0.34 \text{ V}$

A key idea in interpreting these potentials is that a more positive reduction potential means the ion is a stronger oxidizing agent and is more eager to gain electrons (i.e., it is less “stable” in its oxidized form). In contrast, a more negative reduction potential indicates that the species is less likely to gain electrons and is more stable in that oxidation state.

Let’s apply this reasoning:

• For aluminum:

The reaction $\mathrm{Al}^{3+} \to \mathrm{Al}$ has a very negative potential (–1.66 V), meaning that once aluminum is in the +3 state, it does not easily pick up electrons to revert to the metal. This makes $\mathrm{Al}^{3+}$ a very stable oxidation state.

The reaction involving $\mathrm{Al}^{+}$ has a positive potential (+0.55 V), indicating that $\mathrm{Al}^{+}$ is readily reduced to aluminum, and thus is less stable in the +1 state.

• For thallium:

The reaction $\mathrm{Tl}^{3+} \to \mathrm{Tl}$ has a highly positive potential (+1.26 V), so $\mathrm{Tl}^{3+}$ is not very stable; it tends strongly to gain electrons.

In contrast, the reaction involving $\mathrm{Tl}^{+}$ has a negative potential (–0.34 V), which indicates that $\mathrm{Tl}^{+}$ does not easily accept an electron, making it a relatively stable oxidation state.

Now, let’s look at the options:

Option A: “$\mathrm{Tl}^{3+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Because $\mathrm{Tl}^{3+}$ (with +1.26 V) is eager to gain electrons (a strong oxidizing agent), it is less stable compared to $\mathrm{Al}^{3+}$ (with –1.66 V). Thus, Option A is incorrect.

Option B and Option C: “$\mathrm{Al}^{+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Our data indicates that $\mathrm{Al}^{+}$ (with +0.55 V) is less stable (more readily reduced) than $\mathrm{Al}^{3+}$ (with –1.66 V). So, these statements are not correct.

Option D: “$\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.”

• Comparing the two, the reduction reaction for $\mathrm{Tl}^{+}$ has $E^\circ = -0.34\text{ V}$ while that for $\mathrm{Al}^{+}$ has $E^\circ = +0.55\text{ V}.$ The more negative potential for $\mathrm{Tl}^{+}$ shows it is less likely to be reduced to elemental Tl, meaning $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Thus, Option D is the correct statement.

In summary, the correct answer is:

Option D: $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Let's analyze each statement using the given standard reduction potentials:

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

$Cu^{2+} + 2e^- \rightarrow Cu,\quad E^\circ = +0.34\text{ V}$

$Zn^{2+} + 2e^- \rightarrow Zn,\quad E^\circ = -0.76\text{ V}$

For nitrate reducing to nitric oxide (in acidic solution):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Now, let’s consider each statement one by one.

Statement I: “$H^+$ does not oxidise Cu to $Cu^{2+}$”

The possible reaction would be:

$Cu(s) + 2H^+ \rightarrow Cu^{2+} + H_2(g)$

To check spontaneity, write the half-reactions:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,$

(reverse of the given reduction reaction with $E^\circ = -0.34\text{ V}$)

Reduction (hydrogen):

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

The cell potential is:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.0\text{ V} - 0.34\text{ V} = -0.34\text{ V}$

A negative cell potential means the reaction is non-spontaneous.

Thus, Statement I is correct.

Statement II: “Zn reduces $Cu^{2+}$ to Cu”

The reaction is:

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

Write the half-reactions:

Oxidation (zinc):

$Zn(s) \rightarrow Zn^{2+} + 2e^-,$

The oxidation potential for Zn is the reverse of its reduction: $+0.76\text{ V}$.

Reduction (copper):

$Cu^{2+} + 2e^- \rightarrow Cu(s),\quad E^\circ = +0.34\text{ V}$

Combine the potentials:

$E^\circ_{\text{cell}} = 0.34\text{ V} + 0.76\text{ V} = +1.10\text{ V}$

A positive cell potential indicates the reaction is spontaneous.

Thus, Statement II is correct.

Statement III: “$NO_3^-$ oxidises Cu to $Cu^{2+}$”

In reactions with nitric acid (which provides $NO_3^-$ in acidic medium), copper is oxidized. The half-reactions are:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,\quad E^\circ_{\text{ox}} = -0.34\text{ V}$

Reduction (nitrate in acid):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Even though the electrons do not match directly (2 vs. 3 electrons), the overall potential difference can be determined by the difference in the standard electrode potentials:

$E^\circ_{\text{net}} = 0.97\text{ V} - 0.34\text{ V} = +0.63\text{ V}$

A positive net potential shows that the oxidation of Cu by $NO_3^-$ in acidic conditions is thermodynamically favored.

Thus, Statement III is correct.

Since all three statements (I, II, and III) are correct, the correct option is:

Option B: I, II, III

A reducing agent is a substance that donates electrons during a chemical reaction. The more negative the standard reduction potential (given by the half-reaction), the stronger the tendency of the substance to lose electrons—making it a stronger reducing agent.

Let’s compare the given elements by their half-reactions and approximate standard reduction potentials:

For sodium (Na):

$\mathrm{Na^+ + e^- \to Na} \quad E^\circ \approx -2.71\,V$

For rubidium (Rb):

$\mathrm{Rb^+ + e^- \to Rb} \quad E^\circ \approx -2.98\,V$

For magnesium (Mg):

$\mathrm{Mg^{2+} + 2e^- \to Mg} \quad E^\circ \approx -2.36\,V$

For strontium (Sr):

$\mathrm{Sr^{2+} + 2e^- \to Sr} \quad E^\circ \approx -2.89\,V$

Since the strength of a reducing agent corresponds to having a very negative standard reduction potential, we compare the values:

Rb: approximately -2.98 V

Sr: approximately -2.89 V

Na: approximately -2.71 V

Mg: approximately -2.36 V

Rubidium (Rb) has the most negative reduction potential. Therefore, it is the strongest reducing agent among the given elements.

The correct answer is: Option A (Rb).

The standard reduction potentials for the following half-reactions are:

$2 \mathrm{H}^{+} / \mathrm{H}_2$: 0.0 V

$\mathrm{Cu}^{2+} / \mathrm{Cu}$: +0.34 V

$\mathrm{Zn}^{2+} / \mathrm{Zn}$: -0.76 V

$\mathrm{NO}_3^{-}, \mathrm{H}^{-} / \mathrm{NO}$: 0.97 V

Let's consider the reactions:

$\mathrm{Zn} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HNO}_3 \rightarrow$

We are tasked with identifying which of these reactions do not liberate $\mathrm{H}_2(g)$.

Reactions II and III do not produce $\mathrm{H}_2$ because they are not spontaneous. This conclusion is based on their cell potentials ($E_{\text{cell}}^{\circ}$) being negative ($E_{\text{cell}}^{\circ} < 0$).

Calculation of Cell Potentials:

Reaction II: $\mathrm{Cu} + \mathrm{HCl}$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.34 \\ &= -0.34 \, \text{V} \end{align*} $

Reaction III: $\mathrm{Cu} + \mathrm{HNO}_3$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.97 \\ &= -0.97 \, \text{V} \end{align*} $

Since both reactions have negative cell potentials, they are non-spontaneous and thus do not release $\mathrm{H}_2(g)$.

To determine the mass of copper deposited at the cathode during electrolysis, we first need to gather the given data and use the formula for electrolysis. The relevant parameters are:

Current, $ I = 2 \, \text{A} $

Time, $ t = 10 \, \text{minutes} $

Convert the time from minutes to seconds for consistency with the formula:

$ t = 10 \times 60 = 600 \, \text{seconds} $

Next, calculate the total charge $ Q $ passed during electrolysis using the formula:

$ Q = I \times t = 2 \times 600 = 1200 \, \text{C} $

The electrochemical reaction for copper deposition is:

$ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) $

Using the formula for the mass of substance deposited during electrolysis:

$ \text{Mass of copper deposited} = \frac{\text{Molar mass} \times \text{Charge}}{\text{Number of electrons transferred} \times F} $

Given:

Molar mass of Cu = 63 g/mol

Faraday constant, $ F = 96500 \, \text{C/mol} $

Number of electrons transferred for copper, = 2

Substitute these values into the formula:

$ \text{Mass of copper deposited} = \frac{63 \times 1200}{2 \times 96500} = 0.39 \, \text{g} $

Thus, the weight of copper deposited at the cathode is 0.39 grams.

To find the atomic mass of the metal $ M $, given that 2.644 g of metal was deposited with a total charge of 8040 Coulombs, we follow these steps:

Calculate the moles of electrons used:

$ \text{Moles of electrons} = \frac{\text{Total charge}}{\text{Faraday's constant}} = \frac{8040 \, \text{C}}{96500 \, \text{C/mol}} = 0.0833 \, \text{mol} $

Determine the moles of metal deposited:

The reaction $ M\mathrm{F}_2 \rightarrow M^{2+} + 2\mathrm{F}^- $ implies that 2 moles of electrons are needed to deposit 1 mole of metal.

$ \text{Moles of metal} (M) = \frac{\text{Moles of electrons}}{2} = \frac{0.0833}{2} = 0.04165 \, \text{mol} $

Calculate the atomic mass of metal $ M $:

$ \text{Atomic mass of } M = \frac{\text{Mass of metal}}{\text{Number of moles of metal}} = \frac{2.644 \, \text{g}}{0.04165 \, \text{mol}} \approx 63.47 $

Thus, the atomic mass of metal $ M $ is approximately 63.47 u.

In the electrolytic refining process of copper:

The anode is made up of impure copper. During the process, oxidation occurs at the anode, meaning the impure copper dissolves into the electrolyte.

The cathode is made of pure copper. Here, reduction takes place; copper ions gain electrons and deposit as pure copper on the cathode.

Mathematically, the reactions are as follows:

At the anode (oxidation):

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

At the cathode (reduction):

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

Therefore, the correct option is:

Option B: impure copper, pure copper.

The given standard electrode potential is:

$ E^{\circ}_\text{{cell}} = 0.3 \, \text{V} $

According to the Nernst equation:

$ E_\text{{cell}} = E^{\circ}_\text{{cell}} - \frac{2.303RT}{nF} \log \left[\frac{\text{[Cu}^{2+}]}{\text{[M}^{2+}]}\right] \tag{1} $

In this context:

$ n = 2 $ (since it involves the transfer of 2 electrons)

$ E_\text{{cell}} = 0 $ (when the cell voltage is zero)

$\text{[M}^{2+}] = 0.1 \, \text{M} $

$\frac{2.303RT}{F} = 0.06 $

Substituting these values into equation (1) gives:

$ 0.3 = -\frac{0.06}{2} \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Solving for the log term:

$ -\frac{0.6}{0.06} = \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Taking the antilog:

$ 10^{-10} = \frac{\text{[Cu}^{2+}]}{0.1} $

This implies:

$ \text{[Cu}^{2+}] = 10^{-10} \times 10^{-1} = 10^{-11} \, \text{mol/L} $

Therefore, the concentration of $\text{Cu}^{2+}$ required to make the cell potential zero is $\boxed{10^{-11} \, \text{mol/L}}$.

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.

Let's review each statement:

A. During charging of battery, PbSO₄ on anode is converted into PbO₂.

• This statement is incorrect. During charging, PbSO₄ on the anode (lead plate) is converted to Pb, not PbO₂.

B. During charging of battery, PbSO₄ on cathode is converted into PbO₂.

• This statement is correct. During charging, PbSO₄ on the cathode (lead dioxide plate) is converted back to PbO₂.

C. Lead storage battery consists of grid of lead packed with PbO₂ as anode.

• This statement is incorrect. The anode in a lead-acid battery is made of lead (Pb), not lead dioxide (PbO₂). The cathode is made of lead dioxide (PbO₂).

D. Lead storage battery has ∼38% solution of sulphuric acid as an electrolyte.

• This statement is correct. The electrolyte in a lead-acid battery is a solution of about 35-38% sulfuric acid (H₂SO₄) in water.

Therefore, the correct answer is:

B, D only.

The provided reaction is:

$\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

This reaction involves the following half-reactions:

1. Oxidation of hydrogen gas to H+ ions: $\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + e^-$


2. Reduction of AgCl to Ag: $\mathrm{AgCl}(\mathrm{s}) + e^- \rightleftharpoons \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}^{-}(\mathrm{aq})$

Looking at the options provided:

Option A: Doesn't involve H₂ gas, so it can't be correct.

Option B: This includes the necessary elements - H₂, AgCl, and Ag.

Option C: Doesn't involve AgCl, so it can't be correct.

Option D: Also includes the necessary elements - H₂, AgCl, and Ag.

However, looking closely, we can see that Option B represents the galvanic cell for this reaction. The reaction requires the oxidation of H₂ to H⁺, which occurs at the anode. The reaction also requires the reduction of AgCl to Ag and Cl-, which occurs at the cathode.

In Option B, the anode (on the left) is where H₂ is being oxidized to H⁺. The cathode (on the right) is where AgCl is reduced to Ag and Cl^-. The salt bridge or ion exchange component is HCl, which allows for the flow of ions to balance charge in the cell.

Therefore, the reaction occurs in the galvanic cell represented by Option B.

The standard electrode potential (E°) of a metal ion M⁺/M in aqueous solution is calculated under standard conditions and relates to the tendency of the M⁺ ion to gain an electron to form the neutral metal atom, M.

This is indeed a redox (reduction-oxidation) process, and the standard electrode potential is defined for the reduction half-reaction. E° values are based on the energies involved in the processes of ionization and hydration.

However, the standard electrode potential does not depend on the state (solid, liquid, gas) of the atom being ionized. Ionization of a solid metal atom (Option C) would not directly impact the electrode potential as it is not inherently part of the process of reduction at the electrode that the E° values are measuring.

The correct answer should be Option C : Ionisation of a solid metal atom.

In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results. In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO₂. This occurs because KOH is a basic (alkaline) solution and reacts with carbon dioxide (CO₂) from the air to form potassium carbonate (K₂CO₃).

The reaction between KOH and CO₂ is given by :

KOH + CO₂ $ \to $ K₂CO₃

This reaction will change the concentration of KOH in the solution, which in turn will impact the accuracy of the results obtained in volumetric analysis. For example, if the concentration of KOH is lower than it was when the solution was prepared, then more solution will be needed to reach the endpoint in a reaction, and the results will be inaccurate.

That's why it is important to check the concentration of the KOH solution before using it for volumetric analysis, as stated in the assertion (A). And the reason (R) provides the explanation for why the concentration should be checked. Hence, both the assertion and the reason are correct and the correct answer is (D). Both (A) and (R) are correct and (R) is the correct explanation of (A).

The half-cell reaction will be

$ \begin{aligned} \mathrm{Fe}^{3+}(0.02 \mathrm{M})+e^{-} \longrightarrow & \mathrm{Fe}^{2+}(2.0 \mathrm{M}) ; \\ & E^{\circ}{ }_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=0.771 \mathrm{~V} \end{aligned} $

According to Nernst's equation,

$ E=E_{\mathrm{Cell}}^{\circ}-\frac{2303 R T}{n F} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} $

[Number of electrons involved in reduction, $n=1$ ]

$ \begin{gathered} E=0.771 \mathrm{~V}-\frac{2303 R T}{F} \log \frac{2}{0.02} \\ =0.771 \mathrm{~V}-0.059 \log 100 \\ =0.771-0.118=0.653 \mathrm{~V} \end{gathered} $

Hence, reduction potential of half-cell is 0.653 V .