Which of the following statements are true in accordance with Nernst…

The correct answer is Option (3) → (A), (B) and (C) only

The Nernst equation relates the potential of a cell to the concentrations of the ions involved. For a cell reaction in which n electrons are transferred:

    \( \displaystyle E_{cell}=E^{0}_{cell}-\frac{RT}{nF}\ln Q \)

    \( \displaystyle E_{cell}=E^{0}_{cell}-\frac{0.059}{n}\log Q\quad (\text{at } 298\ \mathrm{K}) \)

Here Q is the reaction quotient: product ion concentrations divided by reactant ion concentrations. Pure solids are left out. At equilibrium, E_cell = 0 and Q becomes K_c:

    \( \displaystyle E^{0}_{cell}=\frac{0.059}{n}\log K_c \)

In the given reaction, 2 electrons are transferred, so n = 2 and Q = \( \ce{[Ni^{2+}]} \)/[Ag⁺]².

• (A) E_cell = E⁰_(cell) − RT/2F ln [Ni²⁺] / [Ag⁺]² — Correct. This is the Nernst equation with n = 2 and the correct reaction quotient.

• (B) The electrode potential of the cell is 2.81V. — Correct.     \( \displaystyle E_{cell}=2.97-\frac{0.059}{2}\log\frac{0.2}{(10^{-3})^{2}} \)

      \( \displaystyle =2.97-0.0295\log(2\times10^{5}) \)

      \( \displaystyle =2.97-0.0295\times(0.30+5) \)

      \( \displaystyle =2.97-0.156\approx2.81\ \mathrm{V} \)

• (C) Equilibrium constant Kc can be calculated from the given data. — Correct. E⁰_(cell) and n are both known, so K_c follows from the equilibrium form of the equation:

      \( \displaystyle \log K_c=\frac{2\times2.97}{0.059} \)

• (D) E_cell = E⁰_(cell) − RT/2F ln [Ag⁺]² / [Ni⁺] — Incorrect. The reaction quotient is inverted. The product ion (\( \ce{Ni^{2+}} \)) must be in the numerator and the reactant ion (\( \ce{Ag^{+}} \)) in the denominator.

Note: It seems that there is a mistake in the original NTA paper. In statement (D), the concentration term is printed as \( \ce{[Ni^{+}]} \); it should be \( \ce{[Ni^{2+}]} \), since the cell reaction forms \( \ce{Ni^{2+}} \) ions.

NCERT (2026-27 edition):
Nernst Equation – …