Chemistry · 2019
NEET 2019 · 5 May · Code P1 · Q82
For 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.…
For a cell involving one electron $\displaystyle \mathrm{E}_{\text {cell }}^{\ominus}=0.59 \mathrm{~V}$ at $\displaystyle 298$ K, the equilibrium constant for the cell reaction is :
$\displaystyle \left[\right.$ Given that $\displaystyle \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ at $\displaystyle \left.\mathrm{T}=298 \mathrm{~K}\right]$
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 1.0 \times 10^{10}$
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NEET (UG) 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.