The Nernst equation numericals mcqs class 12 plays a crucial role in electrochemistry and helps calculate the cell potential under non-standard conditions. Understanding Nernst equation numericals mcqs class 12 is vital for Class 12 Chemistry aspirants, as it allows them to solve problems involving electrochemical cells. The Nernst equation allows for the determination of the relationship between the concentrations of the ions involved in a redox reaction and the cell potential.

By incorporating this formula, aspirants can solve problems that require calculating the electrode potential of half-reactions, understanding the effects of concentration, and predicting cell behavior in real-time scenarios.The Nernst equation numericals mcqs class 12  are often focused on determining cell potentials under varying conditions such as different ion concentrations or temperature changes.

For instance, in a problem involving the Daniell cell,in nernst equation numericals mcqs class 12, the Nernst equation helps to determine the potential when the concentration of ions is different from the standard state (1 M concentration for all species). Understanding how to apply the Nernst equation in such numericals is essential for solving many practical and theoretical problems in electrochemistry.

One important concept related to Nernst equation numericals mcqs class 12 is the understanding of the equilibrium constant, KKK, for a redox reaction. The Nernst equation allows us to calculate the equilibrium potential for a reaction based on the concentrations of the products and reactants. If the reaction reaches equilibrium, the cell potential becomes zero, and the Nernst equation simplifies to give the equilibrium constant, which can then be used to predict the direction of the reaction.

In practice, Nernst equation numericals mcqs class 12 may involve finding the cell potential at equilibrium, which is critical in predicting whether a reaction will proceed spontaneously.Another aspect of Nernst equation numericals is the effect of temperature on the cell potential.

The temperature factor in the Nernst equation allows aspirants to calculate the impact of temperature changes on the cell potential. For example, when the temperature increases, the cell potential can either increase or decrease depending on the sign and nature of the reaction. This temperature dependence of the Nernst equation is useful when solving numericals that involve reactions at different temperatures.

Furthermore, Nernst equation numericals mcqs class 12 often require the use of logarithms to solve for the cell potential. Since the Nernst equation involves a logarithmic term, understanding how to handle logarithmic expressions is essential for successfully solving these problems. Aspirants must be comfortable with converting concentrations and understanding how changes in ion concentrations affect the overall potential of the electrochemical cell.

In addition, aspirants may encounter problems where they need to apply the Nernst equation numericals mcqs class 12 to calculate the cell potential when one of the ion concentrations is given, and the other is unknown. This requires a solid grasp of how to manipulate the Nernst equation numerically to isolate and solve for unknown values, such as concentrations of ions at equilibrium.

Mastering this skill is crucial for solving more advanced Nernst equation numericals mcqs class 12.Nernst equation numericals mcqs class 12 also serve as a foundation for understanding real-world applications of electrochemical cells. From determining the electrochemical behavior of batteries to understanding the principles behind galvanic cells, the Nernst equation is a fundamental tool used in many scientific and industrial applications.

Therefore, it is important for aspirants to practice solving these problems regularly to build a strong foundation in electrochemistry.

Nernst Equation Numericals MCQs Class 12

1.  The strongest oxidizing agent is
   a) Fe³⁺
   b) Fe(CN)₆⁻³
   c) [Fe(CN)₆]⁻⁴
   d) Fe²⁺
   Answer: a

2. Zinc is used to protect iron from corrosion because
   a) EredE_{red}Ered​ of Zn < EredE_{red}Ered​ of iron
   b) EoxidationE_{oxidation}Eoxidation​ of Zn < EoxidationE_{oxidation}Eoxidation​ of iron
   c) EoxidationE_{oxidation}Eoxidation​ of Zn = EoxidationE_{oxidation}Eoxidation​ of iron
   d) Zinc is cheaper than iron
   Answer: a

3. Which is the 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

4. Using the data given below, find out the strongest reducing agent?
   ECr2O72−/Cr3+0E^0_{Cr_2O_7^{2-}/{Cr^{3+}}}ECr2​O72−​/Cr3+0​ = 1.33 V, ECl2/Cl−0E^0_{Cl_2/{Cl^-}}ECl2​/Cl−0​ = 1.36 V
   EMnO4−/Mn2+0E^0_{MnO_4^{-}/{Mn^{2+}}}EMnO4−​/Mn2+0​ = 1.51 V, ECr3+/Cr0E^0_{Cr^{3+}/{Cr}} ECr3+/Cr0​ = -0.74 V
   a) Cl⁻
   b) Cr
   c) Cr³⁺
   d) Mn²⁺
   Answer: b

5. EMF of the following cell at 298 K in V is x * 10⁻²:
   Zn | Zn²⁺ (0.1M) || Ag⁺ (0.01M) | Ag
   Given: EZn2+/Zn0E^0_{Zn²⁺/{Zn}}EZn2+/Zn0​ = -0.76 V; EAg+/Ag0E^0_{Ag⁺/{Ag}}EAg+/Ag0​ = +0.80 V; 2.303RTF\frac {2.303RT}{F}F2.303RT​ = 0.059
   a) 14
   b) 140
   c) 150
   d) 147
   Answer: d

6. The element with the 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

7. The electrode potential of M²⁺/M for 3d series elements shows a positive value for:
   a) Fe
   b) Co
   c) Zn
   d) Cu
   Answer: d

8. Given below are two statements:
   Statement I: The E0E^0E0 value of Ce⁴⁺/Ce³⁺ is 1.74V.
   Statement II: Ce is more stable in Ce⁴⁺ state than Ce³⁺ state.
   Choose the most appropriate answer from the options given below:
   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

9. For the cell reaction:
   Cu(s) | Cu²⁺ (aq,0.1M) || Ag⁺ (aq, 0.01M) | Ag(s)
   the cell potential, E1E_1E1​ is 0.3095 V
   For the cell:
   Cu(s) | Cu²⁺ (aq,0.01M) || Ag⁺ (aq, 0.001M) | Ag(s)
   the cell potential is ___ * 10⁻² V (Rounded off to the nearest integer)
   a) 28
   b) 30
   c) 20
   d) 38
   Answer: a

10. For the galvanic cell:
    Zn(s) + Cu²⁺ (0.02 M) → Zn²⁺ (0.04M) + Cu(s),
    EcellE_{cell}Ecell​ = ____ * 10⁻² V. (Nearest integer)
    a) 109
    b) 100
    c) 10
    d) 110
    Answer: a

11. Potassium chlorate is prepared by electrolysis of KCl in a basic solution as shown by the 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)
    a) 1
    b) 2
    c) 3
    d) 4
    Answer: a

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

13. For the cell:
    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

14. Find the emf of the following cell reaction, given:
    ECr3+/Cr2+0E^0_{Cr^{3+}/{Cr^{2+}}}ECr3+/Cr2+0​ = -0.72 V and EFe2+/Fe0=−0.42VE^0_{Fe^{2+}/{Fe}} = -0.42 VEFe2+/Fe0​=−0.42V at 25°C is
    a) 0.3 V
    b) 0.25 V
    c) 1.14 V
    d) 0.309V
    Answer: a

15. Given half-cell potentials:
    ECr3+/Cr0=0.4VE^0_{Cr^{3+}/{Cr}} = 0.4VECr3+/Cr0​=0.4V and ECr3+/Cr0=0.91VE^0_{Cr^{3+}/{Cr}} = 0.91VECr3+/Cr0​=0.91V, find the standard reduction potential of Cr³⁺/Cr
    a) -1.31V
    b) -1.71V
    c) -0.74V
    d) -0.51V
    Answer: c

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

17. The correct order of E0E^0E0 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

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?
    a) 0.011V
    b) 0.022 V
    c) -0.03 V
    d) -0.011V
    Answer: c

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

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

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

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

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

24. Given EFe3+/Fe2+0E^0_{Fe^{3+}/{Fe^{2+}}}EFe3+/Fe2+0​ = +0.76V and EI2/I−0E^0_{I_2/I^-}EI2​/I−0​ = +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

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

26. Which of the following statements is correct for the cell
    Zn | Zn²⁺ || Cu²⁺ | Cu ?
    a) Cu is anode
    b) Cu is oxidising agent.
    c) The cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu
    d) Zn is reducing agent.
    Answer: c

27. Which one of the following has a potential more than zero?
    a) Pt, 12H2(1atm)∣HCl(1M)\frac{1}{2} H_2(1 atm) | HCl(1 M)21​H2​(1atm)∣HCl(1M)
    b) Pt, 12H2(1atm)∣HCl(2M)\frac{1}{2} H_2(1 atm) | HCl(2 M)21​H2​(1atm)∣HCl(2M)
    c) Pt, 12H2(1atm)∣HCl(0.1M)\frac{1}{2} H_2(1 atm) | HCl(0.1 M)21​H2​(1atm)∣HCl(0.1M)
    d) Pt, 12H2(1atm)∣HCl(0.5M)\frac{1}{2} H_2(1 atm) | HCl(0.5 M)21​H2​(1atm)∣HCl(0.5M)
    Answer: b

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

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
    a) -0.935 V
    b) 0.935 V
    c) 0.335 V
    d) -0.335 V
    Answer: a

30. For a cell involving two electron changes, Ecell0E^0_{cell}Ecell0​ = 0.3V at 25°C. The equilibrium constant K of the reaction is
    a) 10⁻¹⁰
    b) 3 * 10⁻²
    c) 10
    d) 10¹⁰
    Answer: d

Conclusion

In conclusion, Nernst equation numericals mcqs class 12  are essential for mastering electrochemistry concepts in Class 12 Chemistry. By applying the Nernst equation numericals mcqs class 12, aspirants can calculate electrode potentials under non-standard conditions, understand the relationship between concentration and potential, and solve problems involving temperature changes and equilibrium constants. Mastery of these nernst equation numericals mcqs class 12 enhances aspirant’s problem-solving skills and prepares them for more advanced studies in electrochemistry and other related fields.

Additonally, consistent practice of Nernst equation numericals mcqs class 12 enables aspirants to develop accuracy in calculations involving concentration cells, equilibrium constants, and temperature effects. Solving Nernst equation numericals mcqs class 12 also helps aspirants identify common mistakes and apply formulas more efficiently under exam pressure. Thus, mastering Nernst equation numericals mcqs class 12 ensures better performances in Class 12 board exams as well as competitive examinations like NEET, JEE and CUET.