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)
Answer & explanation
Correct answer: option 4
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:
Ecell = Cell potential
E°cell = Standard cell potential (at standard conditions, usually at 25°C and 1 atm pressure)
\(E_{cell} = 0.44 − \frac{0.0591}{2}log \frac{10^{-3}}{1^2}\)
\(⇒ E_{cell} = 0.44 − \frac{0.0591}{2} × (-3)\)
\(⇒ E_{cell} = 0.44 + 0.08865\)
\(⇒ E_{cell} = 0.52865\)
\(⇒ E_{cell} ≈ 0.529 V\)