You visited us 1 times! Enjoying our articles? Unlock Full Access!

Oxidising Reagent and Reducing Reagent

Consider the ...

Question

Consider the following redox reaction:
$2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{⊖}$

S_{2} O_{3}^{2 -}

gets reduced to

S_{4} O_{6}^{2 -}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

S_{2} O_{3}^{2 -}

gets oxidised to

S_{4} O_{6}^{2 -}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I_{2}

gets reduced to

I^{⊖}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I_{2}

gets oxidised to

I^{⊖}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
B $I_{2}$ gets reduced to $I^{⊖}$
C $S_{2} O_{3}^{2 -}$ gets oxidised to $S_{4} O_{6}^{2 -}$
Given the following redox reaction:
$2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{⊖}$
$2 S_{2} O_{3}^{2 -} ⟶ S_{4} O_{6}^{2 -} + 2 e^{-}$ (Oxidation)
$I_{2} + 2 e^{-} ⟶ 2 I^{⊖}$ (Reduction)
Hence, $S_{2} O_{3}^{2 -}$ gets oxidised to $S_{4} O_{6}^{2 -}$ while $I_{2}$ gets reduced to $I^{⊖}$ .

Suggest Corrections

Similar questions

I_{2 (a q)} + 2 S_{2} O_{3}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -}

{N a}_{2} S_{2} O_{3}

I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{-}

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (s) ⟶ S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

S_{2} O_{3}^{2 -} (a q) + 2 {B r}_{2} (l) + 5 H_{2} O (l) ⟶ 2 {S O}_{4}^{2 -} (a q) + 4 {B r}^{-} (a q) + 10 H^{+} (a q)

{N a}_{2} S_{2} O_{3}

I_{2}

M_{1}

M_{2}

{N a}_{2} S_{2} O_{3}

I_{2}

2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}

Join BYJU'S Learning Program