Calculate e.m.f. of the following cell at 298 K :
2Cr() + 3Fe2+ (0.1M) → 2Cr3+ (0.01.M) + 3Fe()
E°(Cr3+ | Cr) = -0.74
E°(Fe2+ | Fe) = -0.44V
E°cell=E°cathode -E°anode
= (-0.44) - (-0.74)V
= 0.30 V
