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The problem asks me to balance the following redox equation using the (ion-electron) half-reaction method: 1)...

The problem asks me to balance the following redox equation using the (ion-electron) half-reaction method:

1) MnO41- (aq) + Cl1- (aq) --> Mn2+ (aq) + Cl2 (g) (acidic solution)

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Answer #1

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MnO4- (aq) + Cl- (aq) --> Mn2+ (aq) + Cl2 (g)

Oxidation half reaction,

Cl- (aq) --> Cl2 (g)

Balance Cl atoms : 2Cl- (aq) -->  Cl2 (g)

Balance charge :  2Cl- (aq) -->  Cl2 (g) + 2e- -------(1)

Reduction half reaction :

MnO4- (aq) -----> Mn2+ (aq)

Balance O atoms : MnO4- (aq) -----> Mn2+ (aq) + 4H2O(l)

Balance H atoms :  MnO4- (aq) +8 H+ (aq) -----> Mn2+ (aq) + 4H2O(l)

Balance charge : MnO4- (aq) + 8H+ (aq) +5e- -----> Mn2+ (aq) + 4H2O(l) -----(2)

The total equation is obtained by adding two equations ,[5 x Eqn(1)] + [2xEqn(2)] gives

2MnO4- (aq) + 16H+ (aq) +10e- +10Cl- (aq) --> 5Cl2 (g) + 10e- + 2Mn2+ (aq) + 8H2O(l)

2MnO4- (aq) + 16H+ (aq) +10Cl- (aq) --> 5Cl2 (g) + 2Mn2+ (aq) + 8H2O(l)

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