Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s) | Mg²⁺(0.001M) || Cu²⁺(0.0001M) | Cu(s) (ii) Fe(s) | Fe²⁺(0.001M) || H⁺(1M) | H₂(g)(1bar) | Pt(s) (iii) Sn(s) | Sn²⁺(0.050M) || H⁺(0.020M) | H₂(g) (1bar) | Pt(s) (iv) Pt(s) | Br⁻(0.010M) | Br₂(l) || H⁺(0.030M) | H₂(g) (1bar) | Pt(s).
Nernst equation: E_cell = E°_cell - (0.059/2) log[Mg²⁺]/[Cu²⁺]; E_cell = 2.68 V
Nernst equation: E_cell = E°_cell - (0.059/2) log[Fe²⁺]·P_H₂/[H⁺]²; E_cell = 0.53 V
Nernst equation: E_cell = E°_cell - (0.059/2) log[Sn²⁺]·P_H₂/[H⁺]²; E_cell = 0.08 V
Nernst equation: E_cell = E°_cell - (0.059/2) log P_H₂/[Br⁻]²[H⁺]²; E_cell = -1.30 V
Explanation
The question requires writing the Nernst equation and calculating emf for four electrochemical cells at 298 K. The Nernst equation formula E_cell = E°_cell - (RT/nF)lnQ is provided in the context, which at 298 K becomes E_cell = E°_cell - (0.059/n)logQ. Standard electrode potentials from NCERT Table 3.1 are used: E°(Mg²⁺/Mg) = -2.37V, E°(Cu²⁺/Cu) = +0.34V, E°(Fe²⁺/Fe) = -0.44V, E°(Sn²⁺/Sn) = -0.14V, E°(Br₂/Br⁻) = +1.09V, and E°(H⁺/H₂) = 0V. For each cell, E°_cell = E°_cathode - E°_anode. The reaction quotient Q is written based on the cell reaction with solids and pure liquids having activity = 1.