1. Analysis of the Redox Reaction

The given overall chemical reaction is: $Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$

We decompose this into two half-reactions based on the oxidation state changes [Per the definition of Redox Reactions]:

• Oxidation Half-Reaction (Anode): $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}$ [Loss of electrons]
• Reduction Half-Reaction (Cathode): $2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s)$ [Gain of electrons]

2. Depicting the Galvanic Cell

According to the IUPAC convention for cell notation, the anode is written on the left and the cathode on the right, separated by a salt bridge (denoted by $||$). Phase boundaries are represented by a vertical line ($|$).

Cell Representation: $Zn(s) \mid Zn^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag(s)$

3. Visual Representation of the Galvanic Cell

4. Electrode Reactions

Note: The reduction reaction at the cathode is multiplied by 2 to balance the electrons transferred [Per the Law of Conservation of Mass and Charge].

Final Solution: The galvanic cell is represented as $Zn(s) \mid Zn^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag(s)$. The anodic reaction is $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}$ and the cathodic reaction is $2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s)$.

Step 1: Analyzing the Electrochemical Reduction

The electrolytic reduction of molten calcium chloride ($CaCl_2$) involves the reduction of calcium ions ($Ca^{2+}$) at the cathode to form metallic calcium ($Ca$).

The half-cell reaction at the cathode is:

$Ca^{2+}(l) + 2e^- \rightarrow Ca(s)$

[Per the definition of electrolytic reduction, the charge required to reduce 1 mole of a metal ion is equal to $n \times F$, where $n$ is the valency of the metal and $F$ is Faraday's constant.]

Step 2: Determining Molar Mass and Stoichiometric Requirements

The atomic mass of Calcium ($Ca$) is approximately $40.0 \, g/mol$.

From the balanced half-reaction, it is evident that:

• 1 mole of $Ca$ ($40.0 \, g$) requires 2 moles of electrons ($2 \, F$ of electricity).

Step 3: Calculating Electricity for 20.0 g of Calcium

Using the unitary method, we determine the amount of Faraday ($F$) required to produce $20.0 \, g$ of $Ca$:

$\text{Electricity required} = \left( \frac{2 \, F}{40.0 \, g} \right) \times 20.0 \, g$

[By the Principle of Stoichiometric Equivalence]

$\text{Electricity required} = \frac{2}{40.0} \times 20.0 \, F$

$\text{Electricity required} = 0.05 \times 20.0 \, F = 1.0 \, F$

Step 4: Visualizing the Electrolytic Cell

Final Solution: The amount of electricity required to produce 20.0 g of Calcium from molten $CaCl_2$ is 1.0 Faraday.

Step 1: Identification of Half-Cell Reactions

To determine the feasibility of the reaction between $Ag^{+}(aq)$ and $Cu(s)$, we must define the oxidation and reduction half-reactions. The overall spontaneous reaction occurs if the cell potential ($E^{\circ}_{cell}$) is positive [Per the Nernst Equation and Gibbs Free Energy relationship $\Delta G^{\circ} = -nFE^{\circ}_{cell}$].

• Oxidation (Anode): $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{-}$
• Reduction (Cathode): $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$

Step 2: Standard Electrode Potentials ($E^{\circ}$)

Based on the electrochemical series (standard reduction potentials at $298 \, K$):

Step 3: Calculating Cell Potential ($E^{\circ}_{cell}$)

The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ [By Definition of Electrochemical Cell Potential].

Substituting the values:

$E^{\circ}_{cell} = E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Cu^{2+}/Cu}$

$E^{\circ}_{cell} = 0.80 \, V - 0.34 \, V$

$E^{\circ}_{cell} = +0.46 \, V$

Step 4: Feasibility Analysis

A reaction is considered feasible if the change in Gibbs Free Energy ($\Delta G^{\circ}$) is negative. Since $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, a positive $E^{\circ}_{cell}$ results in a negative $\Delta G^{\circ}$ [Per the Second Law of Thermodynamics regarding spontaneity].

Since $E^{\circ}_{cell} = +0.46 \, V > 0$, the reaction is spontaneous and therefore feasible.

Final Solution: The reaction between $Ag^{+}(aq)$ and $Cu(s)$ is feasible because the calculated standard cell potential ($E^{\circ}_{cell}$) is $+0.46 \, V$, which is greater than zero, indicating a spontaneous process.

Step 1: Identification of Ionic Species in Solution

In an aqueous solution of silver nitrate ($AgNO_{3}$), the substance dissociates completely according to the following equation: $AgNO_{3}(aq) \rightarrow Ag^{+}(aq) + NO_{3}^{-}(aq)$

Since it is an aqueous solution, water ($H_{2}O$) molecules are also present, which can undergo oxidation or reduction [Per the Nernst Theory of Electrolytic Dissociation].

Step 2: Analysis of Potential Electrode Reactions

At the electrodes, the species with higher reduction potential (for cathode) or lower oxidation potential (for anode) will preferentially react. Because silver electrodes are used, the electrodes are active, meaning they participate in the reaction.

At the Cathode (Reduction):

• Reduction of $Ag^{+}$ ions: $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$ ($E^{\circ} = +0.80 \, V$)
• Reduction of $H_{2}O$: $2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq)$ ($E^{\circ} = -0.83 \, V$)

Since the reduction potential of $Ag^{+}$ is significantly higher than that of $H_{2}O$ [Per the Electrochemical Series], $Ag^{+}$ ions are preferentially reduced to metallic silver.

At the Anode (Oxidation):

• Oxidation of $Ag$ electrode: $Ag(s) \rightarrow Ag^{+}(aq) + e^{-}$ ($E^{\circ}_{ox} = -0.80 \, V$)
• Oxidation of $NO_{3}^{-}$ ions: $2NO_{3}^{-} \rightarrow 2NO_{2} + O_{2} + 2e^{-}$ ($E^{\circ}_{ox} \approx -2.0 \, V$)
• Oxidation of $H_{2}O$: $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$ ($E^{\circ}_{ox} = -1.23 \, V$)

In the presence of an active silver anode, the silver metal itself is more easily oxidized than the water or the nitrate ions [Per the Principle of Preferential Discharge]. Therefore, the anode undergoes dissolution.

Visual Representation of the Electrolytic Cell

Step 3: Synthesis of Final Products

As the current flows through the circuit:

1. The silver anode dissolves into the solution: $Ag(s) \rightarrow Ag^{+}(aq) + e^{-}$.
2. The silver ions from the solution (and those produced at the anode) migrate to the cathode.
3. The silver ions at the cathode gain electrons and deposit as metallic silver: $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$.

Effectively, the net result is the transfer of silver from the anode to the cathode.

Final Solution: The product at the cathode is metallic silver ($Ag$), and the product at the anode is the dissolution of the silver electrode (formation of $Ag^{+}$ ions in the solution).

Step 1: Half-Cell Reactions and Standard Electrode Potentials

To determine the standard cell potential ($E^\circ_{cell}$), we identify the oxidation and reduction half-reactions:

• Oxidation (Anode): $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-$ ; $E^\circ_{ox} = -0.77 \text{ V}$
• Reduction (Cathode): $Ag^+(aq) + e^- \rightarrow Ag(s)$ ; $E^\circ_{red} = +0.80 \text{ V}$

[Per standard electrochemical data, $E^\circ_{Fe^{3+}/Fe^{2+}} = 0.77 \text{ V}$ and $E^\circ_{Ag^{+}/Ag} = 0.80 \text{ V}$]

Step 2: Calculation of Standard Cell Potential ($E^\circ_{cell}$)

The standard cell potential is calculated using the formula: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

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

Step 3: Calculation of Standard Gibbs Energy Change ($\Delta_{r}G^\circ$)

The relationship between $\Delta_{r}G^\circ$ and $E^\circ_{cell}$ is defined as: $\Delta_{r}G^\circ = -nFE^\circ_{cell}$

• Number of moles of electrons transferred ($n$) = 1
• Faraday's constant ($F$) $\approx 96500 \text{ C mol}^{-1}$

$\Delta_{r}G^\circ = -(1) \times (96500 \text{ C mol}^{-1}) \times (0.03 \text{ J C}^{-1})$

$\Delta_{r}G^\circ = -2895 \text{ J mol}^{-1} = -2.895 \text{ kJ mol}^{-1}$

Step 4: Calculation of Equilibrium Constant ($K_c$)

Using the Nernst relationship at equilibrium: $\Delta_{r}G^\circ = -RT \ln K_c$ or $E^\circ_{cell} = \frac{0.0591}{n} \log K_c$ (at $298 \text{ K}$):

$0.03 = \frac{0.0591}{1} \log K_c$

$\log K_c = \frac{0.03}{0.0591} \approx 0.5076$

$K_c = \text{antilog}(0.5076) \approx 3.22$

Visual Representation of the Cell

Final Solution: The standard cell potential $E^\circ_{cell}$ is $0.03 \text{ V}$, the standard Gibbs energy change $\Delta_{r}G^\circ$ is $-2.895 \text{ kJ mol}^{-1}$, and the equilibrium constant $K_c$ is approximately $3.22$.

Step 1: Identifying Ionic Species in Solution

In an aqueous dilute solution of sulfuric acid ($H_{2}SO_{4}$), the following dissociation and ionization processes occur:

• Sulfuric acid (strong electrolyte): $H_{2}SO_{4}(aq) \rightarrow 2H^{+}(aq) + SO_{4}^{2-}(aq)$
• Water (weak electrolyte/solvent): $H_{2}O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$

The ions present in the electrolytic cell are: $H^{+}$, $SO_{4}^{2-}$, and $OH^{-}$. [Per the Arrhenius theory of dissociation].

Step 2: Analysis of Cathodic Reactions

At the cathode, reduction occurs. The competing species are $H^{+}$ ions and $H_{2}O$ molecules. Since only $H^{+}$ ions are present (from the acid), the following reduction half-reaction occurs:

$H^{+}(aq) + e^{-} \rightarrow \frac{1}{2}H_{2}(g)$

Product at Cathode: Hydrogen gas ($H_{2}$). [By standard electrode potential convention, $E^{\circ}_{H^{+}/H_{2}} = 0.00 \, V$].

Step 3: Analysis of Anodic Reactions

At the anode, oxidation occurs. The competing species are $SO_{4}^{2-}$ and $OH^{-}$ (from water). The competing half-reactions are:

[Per the principle of selective discharge, the ion/molecule with the lower oxidation potential (or higher reduction potential) is preferred]. Since $E^{\circ}_{ox}$ for water ($+1.23 \, V$) is significantly lower than that of the sulfate ion ($+2.01 \, V$), water is preferentially oxidized.

Step 4: Visualization of the Electrolytic Process

Step 5: Conclusion of Electrolysis Products

The net cell reaction is the electrolysis of water:

$2H_{2}O(l) \xrightarrow{electricity} 2H_{2}(g) + O_{2}(g)$

Final Solution: The product at the cathode is Hydrogen gas ($H_{2}$) and the product at the anode is Oxygen gas ($O_{2}$).

Step 1: Identification of Half-Cell Reactions and Standard Potentials

The given redox reaction is: $2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)$.

• Oxidation Half-Reaction (Anode): $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^-$. The standard reduction potential is $E^\circ_{Cr^{3+}/Cr} = -0.74\text{ V}$.
• Reduction Half-Reaction (Cathode): $Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)$. The standard reduction potential is $E^\circ_{Cd^{2+}/Cd} = -0.40\text{ V}$.

[Per the Nernstian standard convention, $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$]

Step 2: Calculation of Standard Cell Potential ($E^\circ_{cell}$)

$E^\circ_{cell} = E^\circ_{Cd^{2+}/Cd} - E^\circ_{Cr^{3+}/Cr}$

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

Step 3: Calculation of Standard Gibbs Free Energy Change ($\Delta_r G^\circ$)

The total number of electrons transferred ($n$) in the balanced equation is $6$.

The formula for Gibbs energy is $\Delta_r G^\circ = -nFE^\circ_{cell}$, where $F = 96487\text{ C mol}^{-1}$.

$\Delta_r G^\circ = -(6) \times (96487\text{ C mol}^{-1}) \times (0.34\text{ J C}^{-1})$

$\Delta_r G^\circ = -196833.48\text{ J mol}^{-1} \approx -196.83\text{ kJ mol}^{-1}$

Step 4: Calculation of Equilibrium Constant ($K_c$)

The relationship between $\Delta_r G^\circ$ and $K_c$ is defined by: $\Delta_r G^\circ = -RT \ln K_c$ or $\log K_c = \frac{nE^\circ_{cell}}{0.0591}$ at $298\text{ K}$.

$\log K_c = \frac{6 \times 0.34}{0.0591} = \frac{2.04}{0.0591} \approx 34.5178$

$K_c = \text{antilog}(34.5178) = 3.29 \times 10^{34}$

Final Solution: The standard cell potential $E^\circ_{cell} = +0.34\text{ V}$, the standard Gibbs free energy change $\Delta_r G^\circ = -196.83\text{ kJ mol}^{-1}$, and the equilibrium constant $K_c = 3.29 \times 10^{34}$.

Step 1: Identification of Electrolytic Species

In an aqueous solution of silver nitrate ($AgNO_{3}$), the following ions are present:

• From the electrolyte: $Ag^{+}$ and $NO_{3}^{-}$
• From the solvent (water): $H^{+}$ and $OH^{-}$

[Per the principle of electrolytic dissociation in aqueous media].

Step 2: Analysis of Competitive Discharge at the Cathode

At the cathode, reduction occurs. The competing reactions are:

1. $Ag^{+}(aq) + e^{-} \rightarrow Ag(s) \quad (E^{\circ} = +0.80 \, V)$
2. $2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq) \quad (E^{\circ} = -0.83 \, V)$

Since the standard electrode potential ($E^{\circ}$) for the reduction of $Ag^{+}$ is significantly higher (more positive) than that of water, $Ag^{+}$ ions will be preferentially reduced at the cathode [Per the Electrochemical Series hierarchy].

Step 3: Analysis of Competitive Dissolution/Discharge at the Anode

Because the electrodes are specified as Silver (Ag), the anode is active. In cells with active metallic electrodes, the electrode material itself undergoes oxidation if its oxidation potential is higher than that of the anions or water.

The competing reactions are:

1. $Ag(s) \rightarrow Ag^{+}(aq) + e^{-} \quad (E^{\circ}_{ox} = -0.80 \, V)$
2. $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-} \quad (E^{\circ}_{ox} = -1.23 \, V)$
3. $NO_{3}^{-}(aq) \rightarrow \text{No oxidation at these potentials.}$

Since the oxidation potential of silver ($+0.80 \, V$) is greater than that of water ($-1.23 \, V$), the silver anode will dissolve into the solution [Per the Selective Discharge Theory for active electrodes].

Visual Representation of the Electrolytic Cell

Step 4: Conclusion of Products

Final Solution: The product at the cathode is the deposition of metallic silver ($Ag$), and the product at the anode is the dissolution of the silver electrode into the solution as silver ions ($Ag^{+}$).

Given Variables & Standard Electrode Potentials

To determine the feasibility of the reaction between $Fe^{3+}(aq)$ and $I^{-}(aq)$, we must evaluate the standard Gibbs free energy change ($\Delta G^\circ$) or the standard cell potential ($E^\circ_{cell}$). The spontaneous reaction occurs if $E^\circ_{cell} > 0$.

Based on standard electrochemical data (Table 3.1):

• Reduction half-reaction 1: $Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$; $E^\circ_{Fe^{3+}/Fe^{2+}} = +0.77 \text{ V}$
• Reduction half-reaction 2: $I_2(s) + 2e^- \rightarrow 2I^{-}(aq)$; $E^\circ_{I_2/I^-} = +0.54 \text{ V}$

Step 1: Identifying the Anode and Cathode

For a spontaneous reaction to occur, the species with the higher reduction potential must act as the cathode (undergo reduction), and the species with the lower reduction potential must act as the anode (undergo oxidation). [Per the Nernstian electrochemical framework].

Step 2: Constructing the Balanced Redox Reaction

The proposed reaction is the oxidation of iodide ions by iron(III) ions:

Cathode (Reduction): $2Fe^{3+}(aq) + 2e^- \rightarrow 2Fe^{2+}(aq)$

Anode (Oxidation): $2I^{-}(aq) \rightarrow I_2(s) + 2e^-$

Overall Cell Reaction: $2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s)$

Step 3: Calculation of Standard Cell Potential ($E^\circ_{cell}$)

The standard cell potential is calculated using the formula: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ [By the IUPAC convention for electromotive force].

$E^\circ_{cell} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{I_2/I^-}$

$E^\circ_{cell} = 0.77 \text{ V} - 0.54 \text{ V}$

$E^\circ_{cell} = +0.23 \text{ V}$

Step 4: Feasibility Analysis

Since the calculated $E^\circ_{cell}$ is positive ($+0.23 \text{ V}$), the standard Gibbs free energy change $\Delta G^\circ$ is negative, as determined by the relation $\Delta G^\circ = -nFE^\circ_{cell}$ [Per the Thermodynamic Criterion for Spontaneity].

Final Solution: The reaction $2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s)$ is feasible because the standard cell potential $E^\circ_{cell}$ is positive (+0.23 V).

Step 1: Identification of Half-Cell Reactions and Standard Potentials

The overall cell reaction is: $Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$.

We split the reaction into oxidation and reduction half-cells:

• Oxidation (Anode): $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \, V$
• Reduction (Cathode): $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$; $E^{\circ}_{Ag^{+}/Ag} = 0.80 \, V$

[Per the standard electrode potential table values at 298 K]

Step 2: Calculation of Standard Cell Potential ($E^{\circ}_{cell}$)

The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.

$E^{\circ}_{cell} = 0.80 \, V - 0.77 \, V = 0.03 \, V$

Step 3: Calculation of Standard Gibbs Free Energy ($\Delta_{r}G^{\circ}$)

The Gibbs energy change is given by $\Delta_{r}G^{\circ} = -nFE^{\circ}_{cell}$, where $n$ is the number of electrons transferred, $F$ is Faraday's constant ($96487 \, C \, mol^{-1}$), and $E^{\circ}_{cell} = 0.03 \, V$.

Here, $n = 1$.

$\Delta_{r}G^{\circ} = -(1) \times (96487 \, C \, mol^{-1}) \times (0.03 \, V)$

$\Delta_{r}G^{\circ} = -2894.61 \, J \, mol^{-1} \approx -2.89 \, kJ \, mol^{-1}$

[By the fundamental relationship between thermodynamics and electrochemistry]

Step 4: Calculation of Equilibrium Constant ($K_c$)

The relationship between equilibrium constant and cell potential is defined as: $\Delta_{r}G^{\circ} = -RT \ln K_c$, or equivalently, $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_c$ at 298 K.

$0.03 = \frac{0.0591}{1} \log K_c$

$\log K_c = \frac{0.03}{0.0591} \approx 0.5076$

$K_c = \text{antilog}(0.5076) = 10^{0.5076} \approx 3.22$

Visual Representation of the Electrochemical Cell

Final Solution: The standard cell potential is $0.03 \, V$, the standard Gibbs free energy change is $-2.89 \, kJ \, mol^{-1}$, and the equilibrium constant $K_c$ is $3.22$.

Step 1: Theoretical Foundation & Formula Application

To calculate the molar conductivity ($\Lambda_m$) for a given solution, we utilize the following fundamental relationship:

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

Where:

• $\kappa$ = Conductivity in $S \cdot cm^{-1}$
• $c$ = Concentration in $mol \cdot L^{-1}$ (Molarity)
• $1000$ = Conversion factor for $cm^3 \cdot L^{-1}$

Step 2: Tabulation of Calculated Molar Conductivity

Using the provided data points, we calculate $\Lambda_m$ and $\sqrt{c}$ (in $(mol \cdot L^{-1})^{1/2}$):

Step 3: Graphical Visualization and Extrapolation

Step 4: Determination of $\Lambda^0_m$

The relationship is governed by the Debye-Hückel-Onsager equation:

$\Lambda_m = \Lambda^0_m - A\sqrt{c}$

By plotting $\Lambda_m$ vs $\sqrt{c}$ and extrapolating the linear regression to the y-intercept (where $\sqrt{c} = 0$), we obtain the value for the molar conductivity at infinite dilution.

Based on the calculated slope $m \approx -58.05$ and the trend line passing through the data points:

$\Lambda^0_m \approx 124.0 \, S \cdot cm^2 \cdot mol^{-1}$

Final Solution: The calculated molar conductivity values are [123.7, 118.5, 115.75, 111.06, 106.7] $S \cdot cm^2 \cdot mol^{-1}$ respectively. The extrapolated molar conductivity at infinite dilution $\Lambda^0_m$ is 124.0 $S \cdot cm^2 \cdot mol^{-1}$.

Step 1: Identification of Half-Cell Reactions

Based on the cell notation $Mg(s)|Mg^{2+}(0.001M)||Cu^{2+}(0.0001 M)|Cu(s)$, the electrochemical cell consists of an oxidation half-cell (anode) and a reduction half-cell (cathode).

• Anode (Oxidation): $Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-$
• Cathode (Reduction): $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
• Net Cell Reaction: $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$

[Per the convention of electrochemical cell notation, the anode is written on the left and the cathode on the right]. The number of electrons involved in the transfer process is $n = 2$.

Step 2: Determining Standard Cell Potential ($E^\circ_{cell}$ )

The standard reduction potentials for the electrodes are:

• $E^\circ_{Cu^{2+}/Cu} = +0.34 \, V$
• $E^\circ_{Mg^{2+}/Mg} = -2.37 \, V$

The standard cell potential is calculated as:

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

$E^\circ_{cell} = 0.34 \, V - (-2.37 \, V) = 2.71 \, V$

Step 3: Formulating the Nernst Equation

The Nernst equation for the cell at $298 \, K$ is given by:

$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$

[By the Nernst Equation definition, where $Q = \frac{[Product]}{[Reactant]}$ and considering pure solids have unit activity].

Step 4: Calculation of Emf of the Cell

Substituting the given concentrations: $[Mg^{2+}] = 10^{-3} \, M$ and $[Cu^{2+}] = 10^{-4} \, M$

$E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{10^{-3}}{10^{-4}}$

$E_{cell} = 2.71 - 0.02955 \log(10)$

Since $\log(10) = 1$:

$E_{cell} = 2.71 - 0.02955$

$E_{cell} = 2.68045 \, V$

Visual Representation of Electrochemical Cell

Final Solution: The Nernst equation is $E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$ and the emf of the cell is $2.68 \, V$ (approx).

Step 1: Establishing the Theoretical Framework

According to [Faraday’s First Law of Electrolysis], the mass ($m$) of a substance deposited at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte: $m = ZIt = \frac{MIt}{nF}$, where $M$ is the molar mass, $I$ is current in Amperes, $t$ is time in seconds, $n$ is the number of moles of electrons involved, and $F$ is Faraday’s constant ($\approx 96487 \, C \, mol^{-1}$).

Step 2: Calculating Time ($t$) from Silver Deposition

For cell B ($AgNO_{3}$), the cathodic reaction is: $Ag^+ + e^- \rightarrow Ag(s)$. Here, $n = 1$.

• Mass of Silver ($m_{Ag}$) = $1.45 \, g$
• Molar Mass of Silver ($M_{Ag}$) = $107.87 \, g \, mol^{-1}$
• Current ($I$) = $1.5 \, A$

Rearranging the formula for time ($t$):

$t = \frac{m_{Ag} \times n \times F}{M_{Ag} \times I} = \frac{1.45 \, g \times 1 \times 96487 \, C \, mol^{-1}}{107.87 \, g \, mol^{-1} \times 1.5 \, A}$

$t = \frac{139906.15}{161.805} \approx 864.66 \, s$

Step 3: Calculating Mass of Copper Deposited

For cell C ($CuSO_{4}$), the cathodic reaction is: $Cu^{2+} + 2e^- \rightarrow Cu(s)$. Here, $n = 2$.

• Molar Mass of Copper ($M_{Cu}$) = $63.55 \, g \, mol^{-1}$

$m_{Cu} = \frac{M_{Cu} \times I \times t}{n \times F} = \frac{63.55 \times 1.5 \times 864.66}{2 \times 96487}$

$m_{Cu} = \frac{82424.3}{192974} \approx 0.427 \, g$

Step 4: Calculating Mass of Zinc Deposited

For cell A ($ZnSO_{4}$), the cathodic reaction is: $Zn^{2+} + 2e^- \rightarrow Zn(s)$. Here, $n = 2$.

• Molar Mass of Zinc ($M_{Zn}$) = $65.38 \, g \, mol^{-1}$

$m_{Zn} = \frac{M_{Zn} \times I \times t}{n \times F} = \frac{65.38 \times 1.5 \times 864.66}{2 \times 96487}$

$m_{Zn} = \frac{84795.1}{192974} \approx 0.439 \, g$

Electrochemical Cell Schematic

Final Solution: The current flowed for 864.66 seconds. The mass of copper deposited is 0.427 g, and the mass of zinc deposited is 0.439 g.