JEE Main2015ChemistryElectrochemistryActual
At 298 K , the standard reduction potentials are 1 . 51 V for M n O 4 - | M n 2 + , 1 . 36 V for C l 2 | C l - , 1 . 07 V for B r 2 | B r - , 0 . 54 V for I 2 | I - . At pH = 3 , permanganate is expected to oxidize: R T F = 0.059
Options
- AC l - and B r -
- BB r - and I -
- CI - only
- DC l - , B r - and I -
Correct answer
B. B r - and I -
Step-by-step solution
M n O 4 - + 8 H + + 5 e - → M n 2 + + 4 H 2 O   E M n O 4 - M n 2 + = E o - 0.059 5 log ⁡ M n 2 + M n O 4 - H + 8 = 1.51 - 0.059 5 log ⁡ 1 10 - 3 8 (Assuming M n O 4 - = M n 2 + = 1 M ) = 1.51 - 0.059 5 × 24 = 1.51 - 0.28 = 1.23   V E r e d M n O 4 - M n 2 + o = 1.23   V > E r e d B r 2 B r - o > E r e d I 2 I - o   i.e. it will oxidise B r - and I - , only.