Write the Nernst equation and emf of the following cells at 298 k.
Fe(s) | Fe²⁺ (0.001 M) || H⁺ (1M) | H₂ (g) (1 bar) | Pt(s)

Solution

Anode reaction:

Fe (s) \to {Fe}^{2 +} (aq) + 2 e^{-}

Cathode reaction:

2 H^{+} + 2 e^{-} \to H_{2} (g)

Overall cell reaction:

Fe (s) + 2 H^{+} (aq) \to {Fe}^{2 +} (aq) + H_{2} (g) Here, n = 2, {E^{0}}_{cell} = {E^{0}}_{cathode} - {E^{0}}_{anode} = 0 - (- 0.44 V) = + 0.44 V .

The Nernst equation for E_cell at 298 k can be written as:

E_{cell} = {E^{0}}_{cell} - \frac{0.059}{n} \log \frac{[{Fe}^{2 +}]}{[H^{+}]} = 0.44 - \frac{0.059}{n} \log \frac{10^{- 3}}{1} = 0.44 - \frac{0.0591}{2} (- 3)

= 0.44 – 0.0295 (–3) = 0.44 – 0.0885 = 0.5285 V

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