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Electrochemistry

Question

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Calculate the equilibrium constant for the cell reaction:
$4 {Br}^{-} + O_{2} + 4 H^{+} \to 2 {Br}_{2} + 2 H_{2} O$
(Given $E °_{cell} = 0.16$ )

Solution

we have given that
E⁰ = 0.16

\log K_{C} = \frac{nE °_{cell}}{0.0591} E °_{cell} = 0.16 V, n = 4

\log K_{C} = \frac{(4) \times (0.16 V)}{(0.0591 V)} = 10.8 K_{C} = Antilog 10.8 = 6.31 \times 10^{10}

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