NEET 2019 · ChemistryNernst equationPrevious Year Question

For a cell involving one electron $E_{ce l l}^{\circ} = 0.59$ V at $298$ K, the equilibrium constant for the cell reaction is: (Given that $\frac{2.303 R T}{F} = 0.059$ V at $T = 298$ K)

Answer: (C) $1.0 \times 1 0^{10}$ . \textbf{Answer:} (C) $1.0 \times 1 0^{10}$ \textbf{Solution:} The relation between standard cell potential and equilibrium constant is $E_{ce l l}^{\circ} = \frac{0.059}{n} lo g K_{c} .$ With $n = 1$ , $lo g K_{c} = \frac{E _{ce l l}^{\circ}}{0.059} = \frac{0.59}{0.059} = 10.$ Therefore $K_{c} = 1 0^{10} = 1.0 \times 1 0^{10}$ .

A. $1.0 \times 1 0^{2}$
B. $1.0 \times 1 0^{5}$
C. $1.0 \times 1 0^{10}$ ✓
D. $1.0 \times 1 0^{30}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $1.0 \times 1 0^{10}$ \textbf{Solution:} The relation between standard cell potential and equilibrium constant is $E_{ce l l}^{\circ} = \frac{0.059}{n} lo g K_{c} .$ With $n = 1$ , $lo g K_{c} = \frac{E _{ce l l}^{\circ}}{0.059} = \frac{0.59}{0.059} = 10.$ Therefore $K_{c} = 1 0^{10} = 1.0 \times 1 0^{10}$ .

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For a cell involving one electron Ecell∘​=0.59 V at 298 K, the equilibrium constant for the cell reaction is: (Given that F2.303RT​=0.059 V at T=298 K)

Solution & Explanation

More NEET PYQ solutions

For a cell involving one electron $E_{ce l l}^{\circ} = 0.59$ V at $298$ K, the equilibrium constant for the cell reaction is: (Given that $\frac{2.303 R T}{F} = 0.059$ V at $T = 298$ K)