Question

when kmno4 is added dropwise in faintly alkaline solution of iodide ,then it gets oxidized to
​

Answer 1

Answer:

Correct option is

C

IO

3

−

2KMnO

4

+H

2

O+KI→2MnO

2

+2KOH+KIO

3

I+6OH

−

→IO

3

−

+3H

2

O+6e

−

When alkaline KMnO4 is treated with KI, I

−

is oxidised to IO

3

−

.

Answer 2

Answer:

When KMnO₄ is added dropwise in a faintly alkaline solution of iodide then it gets oxidized to iodate.

​Explanation:

• A chemical compound made of inorganic elements is potassium permanganate (KMnO₄). It is sometimes referred to as permanganate of potash or Condy's crystals.
• Due to its high oxidizing power, potassium permanganate can be utilized as an oxidant in a variety of chemical processes.
• When completing a redox reaction with potassium permanganate, the dark purple solution turns colorless and subsequently changes into a brown solution, demonstrating the oxidizing power of the compound.

The above reaction can be performed in an acidic or alkaline medium. One such reaction is with iodine. Which is given as,

$\mathrm{2KMnO_4+H_2O+I^{-}\rightarrow 2MnO_2+2KOH+IO_3^{-}}$           (1)

$\mathrm{I^{-}+6OH^{-}\rightarrow IO_3^{-}+3H_2O+6e^{-}}$         (2)

Where,

KMnO₄= potassium permanganate

H₂O=water

I⁻=iodine ions

MnO₂= Manganese dioxide

KOH=potassium hydroxide

IO₃⁻=iodate

OH⁻=hydroxide ion (representing alkaline medium)

So, when KMnO₄ is added dropwise in a faintly alkaline solution of iodide then it gets oxidized to iodate.

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