1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 T=298 ]

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Q. For a cell involving one electron $E^{\circ}_{cell}$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [ Given that $\frac{2.303\,RT}{F}$= 0.059 V at T=298 ]

NEETNEET 2019Electrochemistry

Solution:

$E_{cell } = E^{\circ}_{cell } - \frac{0.059}{n} \log Q $
(At equilibrium, $Q = K_{eq} $ and $E_{cell} = 0)$
$0 = E^{\circ}_{cell} - \frac{0.059}{1} \log K_{eq}$ (from equation (i))
$\log \; K_{eq} = \frac{E^{\circ}_{cell}}{0.059} = \frac{0.59}{0.059} = 10$
$K_{eq} = 10^{10} = 1 \times 10^{10}$