The electrode potential of the following half cell at 298 K X I X2+(0.0011 M) II Y2+ (0.01 M) I Y is — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Eºcell = o Y | Y2 E  – o X | X2 E  = 0.36 – (–2.36) = 2.72 Ecell = Eºcell – 06 . log ] Y [ ] X [ 2   = 2.72 – 0.03 log            4 10 = 2.72 – 0.03 log (11 × 10–2) = 2.72 – 0.03 [log 11 – 2] = 2.72 – 0.03 [1 – 2] = 2.72 + 0.03 = 2.75 Ans. 275