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2 years ago

Consider the following reaction at 298K.

Chemistry

1 answer:

lakkis [162]2 years ago

8 0

The true statements are;

K < 0
Eocel < 0

<h3>What is a redox reaction?</h3>

We define a redox reaction as one in which a specie is oxidized and another is reduced.

Now;

Eo cell = cell potential = -0.13 V - (+0.34 V) = -0.47 V

n =number of moles of electrons = 2 mole of electrons

K = equilibrium constant

ΔG = change in free energy

Eo cell = 0.0592/n log K

-0.47 = 0.0592/2 log K

log K = -0.47 * 2/0.0592

K = 1.3 * 10^-16

ΔG = -nFEo cell

ΔG = -(2 * 96500 * -0.47)

ΔG = 90.7kJ

Learn more about Ecell:brainly.com/question/10203847

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Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

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Calculating the value of $K_a$ ,

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