Question

Calculate the potential of Iron electrode in which the concentration of Fe2+ ion is 0.01 M. Given \( E^{\circ}_{\text{Fe}^{2+}/\text{Fe}} = -0.45 \, \text{V at 298 K} \)
[Given: log 10 = 1]

Hint:

The Nernst equation allows you to calculate the potential for an electrode based on the concentration of ions. Remember that the standard potential is measured under standard conditions (1 M concentration, 1 atm pressure).

Answer 1

The Nernst equation determines electrode potential under non-standard conditions: \[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{[\text{Fe}^{2+}]}{[\text{Fe}]}\right) \] Variables: - \( E^{\circ} \) represents the standard electrode potential. - \( n \) is the number of electrons transferred; for Fe\textsuperscript{2+} to Fe, \( n = 2 \). - \( [\text{Fe}^{2+}] \) is the concentration of Fe\textsuperscript{2+}, given as 0.01 M. - \( [\text{Fe}] \) is the concentration of solid iron, which is 1 M under standard conditions. Substituting the values: \[ E = -0.45 - \frac{0.0591}{2} \log(\frac{1}{0.01}) \] \[ E = -0.45 - \frac{0.0591}{2} \times (2) = -0.45 + 0.0591 \] \[ E = - 0·509 \, \text{V} \] The iron electrode's potential is approximately \(-0·509\) V.