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Chemistry · PYQ NEET UG 2019(1)
हिन्दीFor a cell involving one electron \( E_{\text{cell}}^\circ = 0.59\text{ V} \) at \( 298\text{ K} \), the equilibrium constant for the cell reaction is _________. [Given that \( \frac{2.303 RT}{F} = 0.059\text{ V} \) at \( T = 298\text{ K} \)]
Correct answer · Option 2
\( 1.0 \times 10^{10} \)
Explanation
Using \( E^\circ = \frac{0.059}{n} \log K_c \), we get \( 0.59 = \frac{0.059}{1} \log K_c \), so \( \log K_c = 10 \), which implies \( K_c = 1.0 \times 10^{10} \).
