Electrode potential calculation mcqs class 12 is an essential concept in physical chemistry, particularly for understanding redox reactions and electrochemical cells. At its core, electrode potential refers to the potential difference between an electrode and its solution, which is a measure of the ability of the electrode to either gain or lose electrons in a redox reaction. In this context, the term electrode potential calculation mcqs class 12 plays a pivotal role in determining the direction of electron flow in electrochemical processes. This article explores the significance and application of electrode potential calculation mcqs class 12 in various electrochemical scenarios.

In electrode potential calculation mcqs class 12, the electrode potential of a half-cell is a measure of its tendency to either gain or lose electrons. The calculation of electrode potential is fundamental when studying electrochemical cells, as it helps predict the cell voltage and the spontaneity of the reactions taking place. To calculate electrode potential, the standard electrode potential values for various half-reactions are used, which are typically given in standard conditions of 25°C, 1 M concentration of ions, and 1 atm pressure for gases. This allows for the electrode potential calculation mcqs class 12 of a half-cell to be compared to other half-cells, providing insights into the electrochemical behavior of various substances.

In a redox reaction, the electrode potential calculation mcqs class 12 determines which half-reaction occurs at the anode and which occurs at the cathode. A positive electrode potential indicates that the half-reaction tends to be reduced, while a negative electrode potential suggests that it is more likely to undergo oxidation. The standard electrode potential of a half-cell is often expressed in volts (V), and when combined with the Nernst equation, it enables the calculation of electrode potentials under non-standard conditions. The electrode potential calculation mcqs class 12 can be used to determine the overall cell potential by subtracting the anode potential from the cathode potential.

A key concept related to electrode potential calculation mcqs class 12 is the relationship between electrode potentials and Gibbs free energy. The Gibbs free energy change (ΔG) for a redox reaction can be linked to the electrode potential through the equation ΔG = -nFE. Here, n is the number of electrons involved in the reaction, F is the Faraday constant, and E is the electrode potential. This equation helps establish a connection between the electrochemical behavior of a system and its thermodynamic properties, providing a more comprehensive understanding of electrochemical reactions. The electrode potential calculation mcqs class 12 can therefore indicate whether a reaction is spontaneous or non-spontaneous, with negative ΔG indicating a spontaneous reaction.

One common application of electrode potential calculation mcqs class 12 is in determining the cell potential of an electrochemical cell. The cell potential is the difference in the electrode potentials of the cathode and anode. For example, in a galvanic cell, the half-reactions at the two electrodes lead to a flow of electrons from the anode to the cathode, and the electrode potential calculation mcqs class 12 helps determine the overall voltage of the cell. The Nernst equation can also be applied in such calculations, allowing for the adjustment of electrode potentials based on the concentration of ions in the solution.

Furthermore, the electrode potential calculation mcqs class 12 is essential when studying the corrosion of metals. Corrosion occurs due to electrochemical reactions between a metal and its environment, and the electrode potential calculation mcqs class 12 of the metal in the presence of certain ions determines the likelihood of corrosion. Metals with more negative electrode potentials are more likely to corrode, as they tend to lose electrons more readily. By understanding the electrode potential calculation mcqs class 12 for different metals, strategies to prevent corrosion, such as the use of sacrificial anodes, can be designed.

Electrode Potential Calculation MCQs Class 12:

1. On the basis of the following E° values, the strongest oxidizing agent is:

   • [Fe(CN)₆]⁻⁴ → [Fe(CN)₆]⁻³ + e⁻; E° = -0.35 V

   • Fe²⁺ → Fe³⁺ + e⁻; E° = -0.77 V

   • a) Fe³⁺

   • b) Fe(CN)₆⁻³

   • c) [Fe(CN)₆]⁻⁴

   • d) Fe²⁺
     Answer: a) Fe³⁺

2. Zinc is used to protect iron from corrosion because:

   • a) Eₛ of Zn < Eₛ of iron

   • b) Eₛ of Zn > Eₛ of iron

   • c) Eₛ of Zn = Eₛ of iron

   • d) Zinc is cheaper than iron
     Answer: a) Eₛ of Zn < Eₛ of iron

3. Which is symbolic representation for the following cell reaction, Mg(s) + Cl₂(g) → Mg²⁺(aq) + 2Cl⁻(aq)?

   • a) Mg | Mg²⁺(aq) (1 M) || Cl⁻(aq) (1 bar) | Pt

   • b) Pt | Cl⁻(aq) (1 M) | Cl₂(g) (1 bar) || Mg²⁺(aq) (1 M) | Pt

   • c) Mg | Mg²⁺(aq) (1 M) || Cl₂(g) (1 bar) | Cl⁻ (1 M) | Pt

   • d) Pt | Cl₂(g) (1 bar) | Cl⁻(aq) (1 M) || Mg²⁺(aq) (1 M) | Mg
     Answer: a) Mg | Mg²⁺(aq) (1 M) || Cl⁻(aq) (1 bar) | Pt

4. Using the data given below, find out the strongest reducing agent?

   • ECr2O72−/Cr3+0=1.33VE^0_{{Cr_2O_7}^{2-}/{Cr}^{3+}} = 1.33 VECr2​O7​2−/Cr3+0​=1.33V

   • ECl2/Cl−0=1.36VE^0_{{Cl_2}/{Cl^-}} = 1.36 VECl2​/Cl−0​=1.36V

   • EMnO4−/Mn2+0=1.51VE^0_{{MnO_4}^{-}/{Mn}^{2+}} = 1.51 VEMnO4​−/Mn2+0​=1.51V

   • ECr3+/Cr0=−0.74VE^0_{{Cr}^{3+}/{Cr}} = -0.74 VECr3+/Cr0​=−0.74V

   • a) Cl⁻

   • b) Cr

   • c) Cr³⁺

   • d) Mn²⁺
     Answer: b) Cr

5. EMF of the following cell at 298 K in V is x * 10⁻².

   • Zn | Zn²⁺ (0.1M) || Ag⁺ (0.01M) | Ag

   • The value of x is ___ (Rounded off to the nearest integer)

   • a) 14

   • b) 140

   • c) 150

   • d) 147
     Answer: d) 147

6. The element with highest standard reduction potential (in Volt) [M²⁺ → M] among the 1st row of transition elements is:

   • a) Ti

   • b) Cr

   • c) Ni

   • d) Cu
     Answer: d) Cu

7. The electrode potential of M²⁺/M for 3d series elements shows positive value for:

   • a) Fe

   • b) Co

   • c) Zn

   • d) Cu
     Answer: d) Cu

8. Given below are two statements.

   • Statement I: The E° value of Ce⁴⁺/Ce³⁺ is 1.74V.

   • Statement II: Ce is more stable in Ce⁴⁺ state than Ce³⁺ state.

   • Choose the most appropriate answer:

   • a) Both statement I and statement II are correct.

   • b) Statement I is incorrect but statement II is correct.

   • c) Both statement I and statement II are incorrect.

   • d) Statement I is correct but statement II is incorrect.
     Answer: d) Statement I is correct but statement II is incorrect.

9. For the cell reaction:

   • Cu(s) | Cu²⁺ (aq,0.1M) || Ag⁺ (aq, 0.01M) | Ag(s)

   • the cell potential, E₁ is 0.3095 V

   • For the cell,

   • Cu(s) | Cu²⁺ (aq,0.01M) || Ag⁺ (aq, 0.001M) | Ag(s)

   • the cell potential – ___ * 10⁻² V (Rounded off to the nearest integer)

   • a) 28

   • b) 30

   • c) 20

   • d) 38
     Answer: a) 28

10. For the galvanic cell:

• Zn(s) + Cu²⁺ (0.02 M) → Zn²⁺ (0.04M) + Cu(s)

• EcellE_{cell}Ecell​ = ____ * 10⁻² V. (Nearest integer)

• ECu/Cu2+0=−0.34VE^0_{{Cu}/{Cu²⁺}} = -0.34 VECu/Cu2+0​=−0.34V; (E^0_{{Zn}/{Zn²⁺}} = +0.76 V)

• a) 109

• b) 100

• c) 10

• d) 110
  Answer: a) 109

11. Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation:

• 6OH⁻ + Cl⁻ → ClO₃⁻ + 3H₂O + 6e⁻

• A current of xA has to be passed for 10h to produce 10.0 g of potassium chlorate. The value of x is ____ (Nearest integer)

• (Molar Mass of KClO₃ = 122.6 g/mol, F = 96500 C)

• a) 1

• b) 2

• c) 3

• d) 4
  Answer: a) 1

12. A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of pH = 9 and passing hydrogen gas around the platinum wire at 1.2 atm pressure. The oxidation potential of such an electrode equals ___ V:

• a) 0.59 V

• b) 0.531 V

• c) 0.69 V

• d) -0.059 V
  Answer: b) 0.531 V

13. For the cell reaction Cu | Cu²⁺ (0.1M) || Cu²⁺ (1.0M) | Cu; the emf of the cell at 25°C is:

• a) 0.059 V

• b) 0.531 V

• c) 0.369 V

• d) 0.029 V
  Answer: d) 0.029 V

14. Find the emf of the following cell reaction, given:

• ECr3+/Cr2+0=−0.72VE^0_{{Cr}^{3+}/{Cr}^{2+}} = -0.72 VECr3+/Cr2+0​=−0.72V and EFe2+/Fe0=−0.42VE^0_{{Fe}^{2+}/{Fe}} = -0.42 VEFe2+/Fe0​=−0.42V at 25°C is Cr | Cr³⁺ (0.1M) || Fe²⁺ (0.1M) | Fe

• a) 0.3 V

• b) 0.25 V

• c) 1.14 V

• d) 0.309V
  Answer: a) 0.3 V

15. Given half-cell potentials ECr3+/Cr2+0=0.4VE^0_{{Cr}³⁺/{Cr}²⁺} = 0.4VECr3+/Cr2+0​=0.4V and (E^0_{{Cr}³⁺/{Cr}} = 0.91V. Find the standard reduction potential of Cr³⁺/Cr:

• a) -1.31V

• b) -1.71V

• c) -0.74V

• d) -0.51V
  Answer: c) -0.74V

16. In the electrolysis of a CuSO₄ solution, how many grams of Cu are plated out on the cathode, in the time that is required to liberate 5.6 L of O₂(g), measured at 1 atm and 273 K, at the anode?

• a) 6.34 g

• b) 4.22 g

• c) 3.17 g

• d) 31.75 g
  Answer: d) 31.75 g

17. The correct order of E° values of the given elements is: (M²⁺ + 2e⁻ → M(s))

• a) Cu²⁺ > Ni²⁺ > Pb²⁺ > Fe²⁺

• b) Ni²⁺ >Fe²⁺ > Cu²⁺ > Pb²⁺

• c) Cu²⁺ > Pb²⁺ > Ni²⁺ > Fe²⁺

• d) Pb²⁺ > Ni²⁺ > Fe²⁺ > Cu²⁺
  Answer: c) Cu²⁺ > Pb²⁺ > Ni²⁺ > Fe²⁺

18. For the reaction:

• Sn(s) | Sn²⁺ (0.10 M) || Pb²⁺ (0.05M) | Pb(s). What will be the approximate EcellE_{cell}Ecell​ when Pb²⁺ is 0.1 M

• (Given ESn2+/Sn0=−0.136VE^0_{{Sn}²⁺/{Sn}} = -0.136 VESn2+/Sn0​=−0.136V; EPb2+/Pb0=−0.126VE^0_{{Pb}²⁺/{Pb}} = -0.126 VEPb2+/Pb0​=−0.126V)

• a) 0.011V

• b) 0.022 V

• c) -0.03 V

• d) -0.011V
  Answer: c) -0.03 V

19. For the reaction:

• 3Sn⁴⁺ + 2Cr → 3Sn²⁺ + 2Cr³⁺, Ecell0E^0_{cell}Ecell0​ is 0.89 V

• Then ΔG° for the reaction is ___

• a) -515.31 kJ/mol

• b) -254.41 kJ/mol

• c) -457.41 kJ/mol

• d) -347.40 kJ/mol
  Answer: a) -515.31 kJ/mol

20. In the Daniell cell, Zn | Zn⁺ || Cu²⁺ | Cu, when an external voltage is applied such that EexternalE_{external}Eexternal​ > EcellE_{cell}Ecell​, current flows from:

• a) Cu to Zn

• b) no current flows

• c) data insufficient

• d) Zn to Cu
  Answer: d) Zn to Cu

21. The equation that is incorrect is:

• a) (Λm0)NaBr−(Λm0)NaCl=(Λm0)KBr−(Λm0)KCl(\Lambda^0_m)_{NaBr} – (\Lambda^0_m)_{NaCl} = (\Lambda^0_m)_{KBr} – (\Lambda^0_m)_{KCl}(Λm0​)NaBr​−(Λm0​)NaCl​=(Λm0​)KBr​−(Λm0​)KCl​

• b) (Λm0)KCl−(Λm0)NaCl=(Λm0)KBr−(Λm0)NaBr(\Lambda^0_m)_{KCl} – (\Lambda^0_m)_{NaCl} = (\Lambda^0_m)_{KBr} – (\Lambda^0_m)_{NaBr}(Λm0​)KCl​−(Λm0​)NaCl​=(Λm0​)KBr​−(Λm0​)NaBr​

• c) (Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBr(\Lambda^0_m)_{NaBr} – (\Lambda^0_m)_{NaI} = (\Lambda^0_m)_{KBr} – (\Lambda^0_m)_{NaBr}(Λm0​)NaBr​−(Λm0​)NaI​=(Λm0​)KBr​−(Λm0​)NaBr​

• d) (Λm0)H2O=(Λm0)HCl+(Λm0)NaOH−(Λm0)NaCl(\Lambda^0_m)_{H_2O} = (\Lambda^0_m)_{HCl} + (\Lambda^0_m)_{NaOH} – (\Lambda^0_m)_{NaCl}(Λm0​)H2​O​=(Λm0​)HCl​+(Λm0​)NaOH​−(Λm0​)NaCl​
  Answer: c) (Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBr(\Lambda^0_m)_{NaBr} – (\Lambda^0_m)_{NaI} = (\Lambda^0_m)_{KBr} – (\Lambda^0_m)_{NaBr}(Λm0​)NaBr​−(Λm0​)NaI​=(Λm0​)KBr​−(Λm0​)NaBr​

22. 108 g of silver (molar mass 108 g mol⁻¹) is deposited at cathode from AgNO₃(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273K and 1 bar pressure from water by the same quantity of electricity is ___

• a) 5.765 L

• b) 4.2 L

• c) 21 L

• d) 2.8 L
  Answer: a) 5.765 L

23. 250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1M AgNO₃ and 0.1M AuCl. The solution was electrolysed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be:

• a) only silver

• b) silver and gold in equal mass proportion

• c) only gold

• d) silver and gold in proportion to their atomic weights
  Answer: c) only gold

24. Given EFe3+/Fe2+0=+0.76VE^0_{{Fe}^{3+}/{Fe}^{2+}} = +0.76VEFe3+/Fe2+0​=+0.76V and (E^0_{{I_2}/{I^-}} = +0.55V. The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is:

• a) 1 * 10⁷

• b) 1 * 10⁹

• c) 3 * 10⁸

• d) 5 * 10¹²
  Answer: a) 1 * 10⁷

25. The correct statement about Cr²⁺ and Mn³⁺ among the following is (Given, atomic numbers of Cr = 24 and Mn = 25):

• a) Cr²⁺ is a reducing agent

• b) Mn³⁺ is a reducing agent

• c) Both Cr²⁺ and Mn³⁺ exhibit d⁴ outer electronic configuration

• d) When Cr²⁺ is used as reducing agent, it attains d⁵ electronic configuration
  Answer: a) Cr²⁺ is a reducing agent

26. Which of the following statements is correct for the cell:

• Zn | Zn²⁺ || Cu²⁺ | Cu?

• a) Cu is anode

• b) Cu is oxidizing agent.

• c) The cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu

• d) Zn is reducing agent.
  Answer: c) The cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu

27. Which one of the following has a potential more than zero?

• a) Pt, 12\frac{1}{2}21​ H₂(1 atm) | HCl(1 M)

• b) Pt, 12\frac{1}{2}21​ H₂(1 atm) | HCl(2 M)

• c) Pt, 12\frac{1}{2}21​ H₂(1 atm) | HCl(0.1 M)

• d) Pt, 12\frac{1}{2}21​ H₂(1 atm) | HCl(0.5 M)
  Answer: b) Pt, 12\frac{1}{2}21​ H₂(1 atm) | HCl(2 M)

28. For which of the following electrolytes the graph of Λₘ against √C gives a negative slope?

• a) Acetic acid

• b) Sodium acetate

• c) Ammonium hydroxide

• d) Water
  Answer: b) Sodium acetate

29. What would be the electrode potential for the given half-cell reaction at pH = 5?

• 2H₂O → O₂ + 4H⁺ + 4e⁻; Ered0E^0_{red}Ered0​ = 1.23V

• (R = 8.314 J mol⁻¹ K⁻¹; Temp = 298 K; oxygen under std. atm. pressure of 1 bar)

• a) -0.935 V

• b) 0.935 V

• c) 0.335 V

• d) -0.335 V
  Answer: a) -0.935 V

30. For a cell involving one electron, Ecell0E^0_{cell}Ecell0​ = 0.59 V at 298 K, the equilibrium constant for cell reaction is:

• (Given that 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059 V at T = 298 K)

• a) 1.0 x 10⁵

• b) 1.0 x 10²

• c) 1.0 x 10⁷

• d) 1.0 x 10¹⁰
  Answer: d) 1.0 x 10¹⁰

Conclusion on Electrode Potential Calculation MCQs Class 12

In addition, electrode potential calculation mcqs class 12 is integral to understanding the electrolysis process. Electrolysis involves the use of an external voltage to drive a non-spontaneous reaction, and by calculating the electrode potential calculation mcqs class 12  of various half-reactions, one can determine the necessary voltage required to initiate the process. This is particularly useful in industries such as electroplating and the extraction of metals from their ores.

The electrode potential calculation mcqs class 12  is not limited to aqueous systems; it also applies to non-aqueous and molten electrolyte systems used in practical electrochemical processes.Practicing electrode potential calculation mcqs class 12 helps aspirants understand how electrode potentials vary under different conditions. Regular exposure to electrode potential calculation mcqs class 12 strengthens the ability to predict the feasibility and directions accurately. This makes electrode potential calculation mcqs class 12 especially important for scoring well in competitive exams like NEET, JEE and CUET.