Question

Write the Nernst equation and also explain the relation between electrode potential \( (E_{cell}) \) and the equilibrium constant \( (K_c) \).

Hint:

Write Ecell = E cell standard minus (0.0591/n) log Q; at equilibrium Ecell = 0 and Q = Kc.

Answer 1

Step 1: Nernst equation. For a general cell reaction
\[ aA + bB \rightarrow cC + dD \]
the Nernst equation gives the cell potential at any concentration:
\[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF}\ln Q \]
where \(E^{\circ}_{cell}\) is the standard cell potential, \(R\) is the gas constant, \(T\) the temperature, \(n\) the number of electrons transferred, \(F\) the Faraday constant, and \(Q\) the reaction quotient.
Step 2: Form at 298 K. Substituting the values of \(R\), \(F\) and converting \(\ln\) to \(\log\):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n}\log Q \]
Step 3: At equilibrium. When the cell reaction reaches equilibrium, the cell can do no more work, so \(E_{cell} = 0\), and the reaction quotient becomes the equilibrium constant, \(Q = K_c\). Substituting:
\[ 0 = E^{\circ}_{cell} - \frac{0.0591}{n}\log K_c \]
Step 4: Relation. Rearranging gives the link between standard cell potential and the equilibrium constant:
\[\boxed{ E^{\circ}_{cell} = \frac{0.0591}{n}\log K_c }\]
A large positive \(E^{\circ}_{cell}\) means a large \(K_c\), i.e. the cell reaction proceeds nearly to completion.