Question

Balance Mno4-+SO2 gives Mn2+ + HSO4- in acidic medium using oxidation number method​

Answer 1

Answer:

$2 \mathrm{MnO}_4^{-(a q)}+6 \mathrm{I}^{-(a q)}+4 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{-(a q)} \rightarrow 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{HSO}_4^{-}(\mathrm{aq})$

Explanation:

The basis for the oxidation number technique is the difference between the oxidation numbers of the oxidising and reducing agents. Redox reactions are split into oxidation and reduction halves for the purposes of the half-reaction technique. The technique to use and choose depends on the individual.

The unbalanced chemical equation is:

$\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{SO}_2(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{HSO}_4^{-}(\mathrm{aq})$

The oxidation half reaction is

$\mathrm{SO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HSO}_4^{-}(\mathrm{aq})+3 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}(\mathrm{aq})$

The reduction half reaction is

$\mathrm{MnO}_4^{-}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{(}(2+)(\mathrm{aq})$

In the reduction half reaction, the oxidation number of Mn changes from $+7$ to +2. Hence, 5 electrons are added to LHS of the reaction.

$\mathrm{MnO}_4^{-}(\mathrm{aq})+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$

Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.

$\mathrm{MnO}_4^{-}(\mathrm{aq})+5 \mathrm{e}^{-}+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$

To balance $\mathrm{O}$ atoms, 4 water molecules are added on RHS.

$\mathrm{MnO}_4^{-}(\mathrm{aq})+5 \mathrm{e}^{-}+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2 .

$5 \mathrm{SO}_2(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 5 \mathrm{HSO}_4^{-}(\mathrm{aq})+15 \mathrm{H}^{+}(\mathrm{aq})+10 \mathrm{e}^{-}(\mathrm{aq})$

$2 \mathrm{MnO}_4^{-}(\mathrm{aq})+10 \mathrm{e}^{-}+16 \mathrm{H}^{+} \text {(aq) } \rightarrow 2 \mathrm{Mn}^{2+} \text { (aq) }$

Two half cell reactions are added to obtain a balanced equation.

$2 \mathrm{MnO}_4^{-}(\mathrm{aq})+5 \mathrm{SO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{Mn}^{2+}(\mathrm{aq})+5 \mathrm{HSO}_4^{-}(\mathrm{aq})$

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