What is the half equation for Fe2+ + MnO4-=Fe3+ +Mn2+ ?

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What is the half equation for Fe2+ + MnO4-=Fe3+ +Mn2+ ?

i need help!!!!!!!!!11What is the half equation for Fe2+ + MnO4-=Fe3+ +Mn2+

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Answer 1 · 2018-04-18T19:19:17

Half equations are exclusively OXIDATION or REDUCTION reactions, in which electrons are introduced as virtual particles...

Explanation:

$Ferrous ion$ $"Ferrous ion"$ is oxidized:

$F e^{2 +} \to F e^{3 +} + e^{-}$ $Fe^(2+) rarr Fe^(3+) + e^(-)$ $(i)$ $(i)$

And $permanganate ion$ $"permanganate ion"$ is reduced:

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O (l)$ $MnO_4^(-)+8H^+ +5e^(-)rarr Mn^(2+) + 4H_2O(l)$ $(i i)$ $(ii)$

For each half-equation charge and mass are balanced ABSOLUTELY, and thus it reflects stoichiometry. TO produce a balanced equation, we adds $(i)$ $(i)$ and $(i i)$ $(ii)$ in such a way as to remove the electrons as virtual particles of convenience...

i.e. $5 \times (i) + (i i)$ $5xx(i)+(ii)$ gives....

$5 F e^{2 +} + M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to 5 F e^{3 +} + M n^{2 +} + 4 H_{2} O (l) + 5 e^{-}$ $5Fe^(2+) +MnO_4^(-)+8H^+ +5e^(-)rarr 5Fe^(3+)+Mn^(2+) + 4H_2O(l) + 5e^(-)$

...and after cancellation....

$5 F e^{2 +} + M n O_{4}^{-} + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} + 4 H_{2} O (l)$ $5Fe^(2+) +MnO_4^(-)+8H^+ rarr 5Fe^(3+)+Mn^(2+) + 4H_2O(l)$

The which is balanced with respect to mass and charge....and note that the colour change of the reaction, deep-purple permanganate to almost colourless $M n^{2 +}$ $Mn^(2+)$ , would give us a visual gauge on the extent of reaction.

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What is the half equation for Fe2+ + MnO4-=Fe3+ +Mn2+ ?

i need help!!!!!!!!!11What is the half equation for Fe2+ + MnO4-=Fe3+ +Mn2+

1 Answer

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