Question

Calculate the emf of the following cell at 298 K : Fe(s) | Fe²⁺ (0.01M) || H⁺(1M) | H₂(g) (1 bar), Pt(s). Given E°Cell = 0.44 V.

A cell and its emf is given below : Pt (s) | H₂ (g, 1 bar) | H⁺ (aq, 1 M) || Cu²⁺ (aq, 1M) | Cu (s). emf of the cell = + 0·34 V. Write the reduction half-reaction at cathode.

How is standard Gibbs energy for a reaction related to equilibrium constant ?

Calculate emf of the given cell : Mg (s) | Mg²⁺ (0·1 M) || Cu²⁺ (1·0 × 10⁻³ M) | Cu (s). Given : E°(Cu²⁺/Cu) = + 0·34 V, E°(Mg²⁺/Mg) = – 2·37 V (log 100 = 2)