A) Gibb's equation
B) Gibbs's-Helmholtz equation
C) Nernst equation
D) van der Waals' equation
Correct Answer: C
Solution :
To determine the electrode potential (E) produced, Nemst gave the following equal: \[E={{E}^{o}}-\frac{RT}{nF}{{\log }_{n}}K\] \[E={{E}^{o}}\] \[{{E}^{o}}=\frac{RT}{nF}{{\log }_{n}}K\] or \[{{E}^{o}}=\frac{2.303RT}{nF}{{\log }_{10}}[{{M}^{n+}}]\] E = electrode potential developed \[{{E}^{o}}\]= standard reduction potential R = gas constant \[(8.3\text{ }J/K/mol)\] T = temperature, F = Faraday constant n = number of electron taking pans in reduction and oxidation
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