Redox Reactions · Question 4

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is . (Nearest integer)

Answer

3

Solution

mol of I- = 0.5 × 1.2 = 0.6 mol

mol of MnO4- = 0.5 × 0.2 = 0.1 mol

In basic medium MnO4- is reduced to MnO2 (Mn: +7 → +4, 3 e⁻):

2MnO4- + 6I- + 4H2O 2MnO2 + 3I2 + 8OH-

0.1 mol MnO4- needs 0.3 mol I- (0.6 mol present I- in excess)

I2 liberated = 32 0.1 = 0.15 mol

Titration: I2 + 2S2O32- 2I- + S4O62-

mol of S2O32- = 2 × 0.15 = 0.3 mol

V = 0.30.1 = 3 L

Answer: 3