How do we represent the reduction of $M n O_{2}$ $MnO_2$ by chloride anion to give $M n^{2 +}$ $Mn^(2+)$ and chlorine gas?

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Answer 1 · 2016-10-30T16:15:39

This is a redox reaction: manganese is reduced......

$M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O$ $MnO_2 +4H^+ + 2e^(-) rarr Mn^(2+) + 2H_2O$ $(i)$ $(i)$

...........and chloride ion is oxidized.

$C l^{-} \to \frac{1}{2} C l_{2} (g) + e^{-}$ $Cl^(-) rarr 1/2Cl_2(g) + e^-$ $(i i)$ $(ii)$

$(i) + (i i) \times 2 :$ $(i) + (ii)xx2:$ (we wish to remove electrons from the final redox equation.

$M n O_{2} + 4 H^{+} + 2 C l^{-} \to M n^{2 +} + C l_{2} (g) + 2 H_{2} O$ $MnO_2 +4H^+ + 2Cl^(-) rarr Mn^(2+) + Cl_2(g) + 2H_2O$

How do we represent the reduction of MnO_2MnO2MnO_2 by chloride anion to give Mn^(2+)Mn2+Mn^(2+) and chlorine gas?