What is the EMF of the following cell at 298 K?

Fe(s) | Fe²⁺ (0.001 M) || H⁺ (1M) | H₂(g) (1 bar), Pt(s) (Given E°_cell = +0.44V)

0.529 V

The correct answer is option 4. 0.529 V.

The given cell can be written in reaction form as:

\(Fe + 2H^+ \longrightarrow Fe^{2+} + H_2 (n = 2)\)

According to the Nernst equation

\(E_{cell} = E^0_{cell} − \frac{0.0591}{2}log \frac{[Fe^{2+}]}{[H^{+}]^2}\)

Where:

E_cell = Cell potential

E°_cell = Standard cell potential (at standard conditions, usually at 25°C and 1 atm pressure)

\(E_{cell} = 0.44 − \frac{0.0591}{2}log \frac{10^{-3}}{1^2}\)

\(⇒ E_{cell} = 0.44 − \frac{0.0591}{2} × (-3)\)

\(⇒ E_{cell} = 0.44 + 0.08865\)

\(⇒ E_{cell} = 0.52865\)

\(⇒ E_{cell} ≈ 0.529 V\)