Find the emf of the cell reaction Ni(s) + 2Ag+(0.001 M) -> Ni^2+(0.001 M) + 2Ag(s) (Given E^ cell = 1.05 V, 2.303RT/F = 0.059 at 298 K)

Correct answer: (C) 0.9615 V. Answer: (C) 0.9615 V Solution: The reaction transfers n = 2 electrons. Reaction quotient: Q = [Ni^2+]/[Ag+]^2 = 0.001/(0.001)^2 = 10-310-6 = 103 Applying the Nernst equation: Ecell = E^ cell - 0.059/n Q = 1.05 - 0.059/2 (103) Ecell = 1.05 - 0.0295 × 3 = 1.05 - 0.0885 = 0.9615 V

NEET 2022 · ChemistryNernst equationPrevious Year Question

Find the emf of the cell reaction $Ni (s) + 2 Ag X^{+} (0.001 M) Ni X^{2 +} (0.001 M) + 2 Ag (s)$ (Given $E_{ce l l}^{\circ} = 1.05 V$ , $\frac{2.303 R T}{F} = 0.059$ at $298 K$ )

Answer: (C) $0.9615 V$ . \textbf{Answer:} (C) $0.9615 V$ \textbf{Solution:} The reaction transfers $n = 2$ electrons.

A. $1.0385 V$
B. $1.385 V$
C. $0.9615 V$ ✓
D. $1.05 V$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $0.9615 V$ \textbf{Solution:} The reaction transfers $n = 2$ electrons. Reaction quotient: $Q = \frac{[ Ni X ^{2 +} ]}{[ Ag X ^{+} ] ^{2}} = \frac{0.001}{( 0.001 ) ^{2}} = \frac{1 0 ^{- 3}}{1 0 ^{- 6}} = 1 0^{3}$ Applying the Nernst equation: $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g Q = 1.05 - \frac{0.059}{2} lo g (1 0^{3})$ $E_{ce l l} = 1.05 - 0.0295 \times 3 = 1.05 - 0.0885 = 0.9615 V$

🎯

43,000+ questions in हिंदी & English · one-tap toggle

Practice NEET 2022 Chemistry — free, with instant solutions

43,000+ NEET questions solved step-by-step, chapter-wise, in the MedicNEET app.

Open →📚Practice all NEET Chemistry PYQs chapter-wise →💡Don't get it? Learn the Electrochemistry concepts →

← Chemistry Q23 Chemistry Q25 →

More NEET PYQ solutions

ChemCalculate the emf of the half cell given below: Pt(s) | H2(g, 2 atm) | HCl(aq, 0.02 M) E^ H+/H2 = 0 V (Given: 2.303RT/F = 0.059, 2 = 0.3010)ChemFor a cell involving one electron E^ cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is: (Given that 2.303,RT/F = 0.059 V at T = 298 K)ChemIn the electrochemical (Daniell) cell Zn | ZnSO4(0.01 M) || CuSO4(1.0 M) | Cu, the emf is E_1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 to 0.01 M, the emf changes to E_2. Which one of the following is the correct relationship between E_1 and E_2? (Given RT/F = 0.059)ChemThe pressure of H2 required to make the potential of the H2 electrode zero in pure water at 298 K is:ChemThe correct order of decreasing acidity of the following aliphatic acids is:ChemWhich one of the following reactions does NOT belong to Lassaigne's test?ChemHow many products (including stereoisomers) are expected from the monochlorination of 2-methylbutane, (CH3)2CH-CH2-CH3?ChemThe total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

More: All NEET 2022 Chemistry PYQs · NEET Chemistry PYQs · All NEET PYQs · Previous year papers

Find the emf of the cell reaction Ni(s)+2AgX+(0.001M)​NiX2+(0.001M)+2Ag(s) (Given Ecell∘​=1.05 V, F2.303RT​=0.059 at 298 K)

Solution & Explanation

More NEET PYQ solutions

Find the emf of the cell reaction $Ni (s) + 2 Ag X^{+} (0.001 M) Ni X^{2 +} (0.001 M) + 2 Ag (s)$ (Given $E_{ce l l}^{\circ} = 1.05 V$ , $\frac{2.303 R T}{F} = 0.059$ at $298 K$ )