What is the balanced net ionic equation for the reaction in acidic solution of $Fe^(2+)$ and $KMnO_4$ ?

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What is the balanced net ionic equation for the reaction in acidic solution of $Fe^(2+)$ and $KMnO_4$ ?

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Answer 1 · 2017-11-19T00:21:32

$5Fe^(+2)$ + $MnO_4^(-) + 8H^+$ => $Mn^(+2) + 5Fe^(+3) + 4H_2O$

Explanation:

Potassium is a spectator ion so assume the following reaction ...

$Fe^(+2) + MnO_4^-$ => $Mn^(+2) + Fe^(+3)$

Oxidation Half Rxn: $Fe^(+2)$ => $Fe^(+3) + e^-$
Reduction Half Rxn: $MnO_4^-$ + $5e^-$ => $Mn^(+2)$

Balancing with respect to charge transfer ...
$5Fe^(+2)$ => $5Fe^(+3) + 5e^-$
$MnO_4^-$ + $5e^ -$ => $Mn^(+2)$

Balancing with respect to mass ... (Rxn in acidic medium, add one water for each oxygen needed to the side needing oxygen and follow with the appropriate number of $H^+$ needed to the side needing hydrogen.

$5Fe^(+2)$ + $MnO_4^(-) + 8H^+$ => $Mn^(+2) + 5Fe^(+3) + 4H_2O$

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What is the balanced net ionic equation for the reaction in acidic solution of $Fe^(2+)$ and $KMnO_4$ ?

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What is the balanced net ionic equation for the reaction in acidic solution of Fe^(2+)Fe^(2+) and KMnO_4KMnO_4?

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What is the balanced net ionic equation for the reaction in acidic solution of $Fe^(2+)$ and $KMnO_4$ ?