back to directory
NEET CHEMISTRYElectrochemistryMedium

Question

The Gibb's energy for the decomposition of Al2O3Al_2O_3 at 500C500^{\circ}\text{C} is as follows: 23Al2O343Al+O2 ; ΔrG=+960 kJ mol1\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2 \ ; \ \Delta_rG = + 960 \text{ kJ mol}^{-1} The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3Al_2O_3) at 500C500^{\circ}\text{C} is at least:

A

3.0 V

B

2.5 V

C

5.0 V

D

4.5 V

Step-by-Step Solution

The given reaction for the decomposition of Al2O3Al_2O_3 is: 23Al2O343Al+O2\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2 In this reaction, 43\frac{4}{3} moles of Al3+Al^{3+} ions are reduced to Al atoms. The number of electrons involved (nn) can be calculated from either the reduction of Al or oxidation of O: For Al: n=43×3=4n = \frac{4}{3} \times 3 = 4 moles of electrons. For O: 2O2O2+4e2O^{2-} \rightarrow O_2 + 4e^-, so n=4n = 4. We know the relation between standard Gibbs energy and minimum external potential (EE) required for electrolysis: ΔrG=nFE\Delta_rG = nFE Given ΔrG=+960 kJ mol1=960×103 J mol1\Delta_rG = +960 \text{ kJ mol}^{-1} = 960 \times 10^3 \text{ J mol}^{-1} and F96500 C mol1F \approx 96500 \text{ C mol}^{-1}. 960×103=4×96500×E960 \times 10^3 = 4 \times 96500 \times E E=960×1034×96500=960000386000=2.487 V2.5 VE = \frac{960 \times 10^3}{4 \times 96500} = \frac{960000}{386000} = 2.487 \text{ V} \approx 2.5 \text{ V} Hence, the minimum potential difference needed for the electrolytic reduction is 2.5 V2.5 \text{ V}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistryenergydecompositioncirctextcfollowsfracalo

More Electrochemistry Questions

View all

A cell reaction becomes spontaneous when:

A.$\Delta G^{\circ}$ is negative
B.$\Delta G^{\circ}$ is positive
C.$E^{\circ}_{Red}$ is positive
D.$E^{\circ}_{Red}$ is negative
EasySolve

In a typical fuel cell, the reactants (R) and products (P) are:

A.R = $H_2(g)$, $O_2(g)$; P = $H_2O_2(l)$
B.R = $H_2(g)$, $O_2(g)$; P = $H_2O(l)$
C.R = $H_2(g)$, $O_2(g)$, $Cl_2(g)$; P = $HClO_4(aq)$
D.R = $H_2(g)$, $N_2(g)$; P = $NH_3(aq)$
EasySolve

The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change in free energy will be the function of:

A.$\ln(C_1+C_2)$
B.$\ln(C_2/C_1)$
C.$\ln C_2$
D.$\ln C_1$
MediumSolve

The efficiency of a fuel cell is given by:

A.$\Delta H/\Delta G$
B.$\Delta G/\Delta S$
C.$\Delta G/\Delta H$
D.$\Delta S/\Delta G$
EasySolve

The number of Faradays (F) required to produce $20 \text{ g}$ of calcium from molten $\text{CaCl}_2$ (Atomic mass of Ca = $40 \text{ g mol}^{-1}$) is:

A.2
B.3
C.4
D.1
MediumSolve

For a cell involving one electron $E^\ominus_{\text{cell}} = 0.59 \text{ V}$ at $298 \text{ K}$. The equilibrium constant for the cell reaction is: [Given that $\frac{2.303 RT}{F} = 0.059 \text{ V}$ at $T = 298 \text{ K}$]

A.$1.0 \times 10^{30}$
B.$1.0 \times 10^2$
C.$1.0 \times 10^5$
D.$1.0 \times 10^{10}$
EasySolve

The value of $E^{\circ}_{cell}$ for the following reaction is: $Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$ (Given, equilibrium constant is $10^6$)

A.0.17
B.0.01
C.0.05
D.1.77
MediumSolve

The three cells with their $E^{\circ}_{cell}$ values are given below: 1. $Fe|Fe^{2+}||Fe^{3+}|Fe \ ; \ 0.404\text{ V}$ 2. $Fe|Fe^{2+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 1.211\text{ V}$ 3. $Fe|Fe^{3+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 0.807\text{ V}$ The standard Gibbs free energy change values for three cells are, respectively: (F represents the charge on $1\text{ mole}$ of electrons.)

A.–1.212 F, –1.211 F, –0.807 F
B.+2.424 F, +2.422 F, +2.421 F
C.–0.808 F, –2.422 F, –2.421 F
D.–2.424 F, –2.422 F, –2.421 F
HardSolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →