Question

iodid ion is oxidized to hypoiodite ion by permanganate what is the ratio of hydroxide ions to iodide ions in balanced equation

Answer 1

Answer: The ratio of number of hydroxide ions to iodide ions is 2 : 1.

Step-by-step explanation:

The chemical equation for the oxidation of iodide ion is oxidized to hypoiodite ion by permanganate follows:

`2MnO_4^-+I^-+H_2O\rightarrow 2MnO_2(s)+IO_3^(-)+2OH^(-)`

According to mole concept:

1 mole of a substance contains
`6.022* 10^(23)` number of particles

Moles of iodide ions = 1 mole

Number of iodide ions =
`(1* 6.022* 10^(23)`

Moles of hydroxide ions = 2 mole

Number of hydroxide ions =
`(2* 6.022* 10^(23)`

Taking the ratio of number of hydroxide ions and number of iodide ions are as follows:

`\frac{\text{Number of hydroxide ions}}{\text{Number of iodide ions}}=((2* 6.022* 10^(23)))/((1* 6.022* 10^(23)))\\\\\frac{\text{Number of hydroxide ions}}{\text{Number of iodide ions}}=(2)/(1)`

Hence, the ratio of number of hydroxide ions to iodide ions is 2 : 1.