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in Chemistry by (122k points)
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In Nernst equation, the constant 0.0592 at 298 K represents the value of
A. `(RT)/(nF)`
B. `(RT)/(F)`
C. `(2.303 RT)/(nF)`
D. `(2.303RT)/(F)`

2 Answers

+1 vote
by (66.4k points)
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Best answer

Correct option is: (D) \(\frac{2.303 \ RT}{F}\)

According to Nernst  equation

E = \(E_0 - \frac {RT}{nF}\) ln Q

or E = \(E_0 - \frac {8.314 \times 298 \times 2.303}{n\times 96500}\) log Q.

E = \(E_0 - \frac {0.0592}{n}\) log Q.

Therefore, \(\frac{2.303 \ RT}{F}\) = 0.0592

0 votes
by (121k points)
Correct Answer - B::C::D

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