Problem 66 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The Edison storage cells is represented as \(\mathrm{Fe}(\mathrm{s})|\mathrm{FeO}(\mathrm{s})| \mathrm{KOH}(\mathrm{aq})\left|\mathrm{Ni}_{2} \mathrm{O}_{3}(\mathrm{~s})\right| \mathrm{Ni}(\mathrm{s})\) The half-cell reactions are : \(\mathrm{Ni}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\rightleftharpoons 2 \mathrm{e}^{-} 2 \mathrm{NiO}_{(s)}+2 \mathrm{OH}^{-}\) \(E^{\circ}=+0.40 \mathrm{~V}\) \(\mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}(\mathrm{s})+2 \mathrm{OH}^{-} ; E^{0}=-0.87 \mathrm{~V}\) (i) What is the cell reaction? (ii) What is the cell e.m.f ? How does it depend on the concentration of \(\mathrm{KOH}\) ? (iii) What is the maximum amount of electrical energy that can be obtained from one mole of \(\mathrm{Ni}_{2} \mathrm{O}_{3} ?\)

Step-by-Step Solution

Answer

(i) \(\mathrm{Ni}_{2}\mathrm{O}_{3}(s) + \mathrm{FeO}(s) \rightarrow 2 \mathrm{NiO}_{(s)} + \mathrm{Fe}(s) \). (ii) Cell e.m.f. is 1.27 V; e.m.f. increases with KOH concentration. (iii) Maximum energy is 245.8 kJ/mol.

Step 1: Write the Overall Cell Reaction

The overall cell reaction in a galvanic cell is the combination of the two half-reactions, where the electrons cancel out. First, write the two half-cell reactions:1. Reduction at the cathode: \( \mathrm{Ni}_{2}\mathrm{O}_{3}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{NiO}_{(s)} + 2 \mathrm{OH}^{-} \)2. Oxidation at the anode: \( \mathrm{FeO}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(1) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) + 2 \mathrm{OH}^{-} \)Combine these reactions. Since both have 2 electrons, we can directly combine them:\( \mathrm{Ni}_{2}\mathrm{O}_{3}(\mathrm{s}) + \mathrm{FeO}(\mathrm{s}) \rightarrow 2 \mathrm{NiO}_{(s)} + \mathrm{Fe}(\mathrm{s}) \)

Step 2: Calculate the Cell EMF

The cell electromotive force (e.m.f) is calculated using the standard electrode potentials:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]Substituting values:\[ E^{\circ}_{\text{cell}} = 0.40 \text{ V} - (-0.87 \text{ V}) = 0.40 \text{ V} + 0.87 \text{ V} = 1.27 \text{ V} \]Since KOH is in the electrolyte, changes in its concentration would affect the reaction quotient and thereby the cell potential, according to the Nernst equation.

Step 3: Apply Nernst Equation (Concentration Effect)

The Nernst equation can be used to find how the cell e.m.f. changes with different concentrations of KOH:\[ E = E^{\circ} - \frac{RT}{nF} \ln(Q) \]Here, \(E^{\circ}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of moles of electrons exchanged (2 in this case), \(F\) is Faraday's constant, and \(Q\) is the reaction quotient. The concentration of OH⁻, which comes from KOH, will affect the value of \(Q\). If the concentration of KOH increases, the cell potential typically increases.

Step 4: Calculate Maximum Electrical Energy

The maximum electrical energy obtainable from the cell can be calculated using the formula:\[ \text{Work} = nFE^{\circ}_{\text{cell}} \]Where:- \(n = 2\) moles of electrons,- \(F = 96485\, \text{C/mol} \) (Faraday's constant),- \(E^{\circ}_{\text{cell}} = 1.27\, \text{V} \).Thus:\[ \text{Work} = 2 \times 96485 \times 1.27 = 245,779.9\, \text{J/mol} \] or approximately 245.8 kJ/mol.

Key Concepts

Electrochemical Cell Reactions

An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. These cells consist of two half-cells, each containing a different chemical reaction. In the case of Edison storage cells, these reactions occur between the materials in the cell. For instance, at the cathode, the reaction involves the reduction of Nickel(III) oxide with water, forming Nickel(II) oxide and hydroxide ions. Meanwhile, at the anode, ferric oxide is reduced back to iron with water, releasing hydroxide ions.

The beauty of these cell reactions lies in their ability to transfer electrons. This electron flow is what provides electrical energy. Combining these reactions, as shown in the Edison storage cell example, gives us the overall cell reaction where electrons are balanced and transfer from one half-cell to the other. The overall reaction combines the oxidation of iron compounds and the reduction of nickel oxides, producing a net equation of:

• Nickel oxide production: \[ \mathrm{Ni_2O_3(s)} + \mathrm{FeO(s)} \rightarrow 2 \mathrm{NiO(s)} + \mathrm{Fe(s)} \]

Understanding these reactions is crucial for grasping how the electrochemical cell converts chemical energy to electrical energy.

Cell Electromotive Force (EMF)

Cell Electromotive Force (EMF) is the potential difference between two electrodes of an electrochemical cell. This force drives the flow of electrons through the circuit, converting chemical energy into electrical energy. Calculating the EMF involves the standard reduction potentials of the cathode and the anode.

The Edison storage cell has a reduction potential for the cathode reaction of +0.40 V and for the anode reaction of -0.87 V. The overall EMF of the cell is given by the difference:

• \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
• Substituting in the given values leads to:
  \[ 1.27 \text{ V} \]

This calculated EMF indicates how much energy the cell can provide per unit charge. It's important to understand that changes in concentration of the electrolyte, such as KOH, can affect the reaction quotient, thereby influencing the EMF. The Nernst equation will help in understanding these variations.

Nernst Equation

The Nernst equation provides a quantitative relationship between the concentration of the reacting species and the cell potential. It is particularly useful when considering variations from the standard state, allowing for the calculation of the actual cell potential based on the reaction conditions. In our scenario, it's vital for understanding how KOH concentration impacts the cell's potential.

This equation is expressed as:

• \[ E = E^{\circ} - \frac{RT}{nF} \ln(Q) \]

• Where:
  
  • \(E^{\circ}\) is the standard cell potential
  • \(R\) is the gas constant (8.314 J/mol K)
  • \(T\) is the temperature in Kelvin
  • \(n\) is the number of moles of electrons exchanged (2 in this case)
  • \(F\) is Faraday's constant (96485 C/mol)
  • \(Q\) is the reaction quotient, involving the concentrations of the ionic species.

By affecting the concentration of OH⁻ ions, the concentration of KOH influences the value of \(Q\). This, in turn, adjusts the cell potential \(E\). Typically, as the concentration of the KOH increases, the cell potential increases, reflecting how more reactants can drive the cell reaction forward, thus enhancing electrical output.