At what pH, given half cell MnO (0.1M) | Mn2+ (0.001 M) will have electrode potential of 1.282 V?___ — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

MnO4– + 8H+ + 5e– Mn2+ + 4H2O 10–1 M 10–3 M ERP = E0RP – 4 ] H ][ MnO [ ] Mn [ log 059 .    1.282 = 1.54 – log 059 . 8 3 ] H [ 10     – 0.258 = – 06 . [log 10–2 – 8 log [H+]] – 0.258 = – 06 . [–2 + 8 pH] – 21.5 = 2 –8 pH + 5 . = pH pH = 2.9375  3