2 MnO₄⁻+bI⁻+cH₂ O → x I₂+y MnO₂+z O̅ H
If the above equation is balanced with integer coefficients, the value of z is ________.

Solution

Chemistry ma’am · 4 steps

Balance the electrons first, then let the charge tell you how many OH⁻ you need.

1. Track oxidation states. Mn goes from +7 in MnO₄⁻ to +4 in MnO₂, so each Mn gains 3 electrons. Iodine goes from -1 in I⁻ to 0 in I₂, so each I loses 1 electron.
2. Match the electrons. Two MnO₄⁻ gain 2 × 3 = 6 electrons, so 6 I⁻ must lose them. That gives b = 6, x = 3 and y = 2.
3. Balance the charge. The left side carries 2(-1) + 6(-1) = -8. On the right only OH⁻ is charged, so it needs 8 of them: z = 8. Hydrogen then needs 2c = 8, so c = 4.

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

4. So. Oxygen checks out: 8 + 4 = 12 on the left and 4 + 8 = 12 on the right. The value of z is 8.

The idea to keep. In basic medium, balance the electron transfer first, then use OH⁻ to fix the charge and H₂O to fix hydrogen. Check oxygen last.

Answer from the NTA final answer key. Solution by PrepWiser.