If we balanced the following reaction , what is the net ionic charge on the right side of the equation?
$. . . H^{+} + . . . M n O_{4}^{-} + . . . F e^{2 +} \to$
$. . . M n^{2 +} + . . . F e^{3 +} + . . . H_{2} O$

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The net ionic charge on either side must be zero.

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Solution

The correct option is C 17
The unbalanced equation is $H^{+} + M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +} + H_{2} O$
The oxidation number of iron increases from +2 to +3. Increase in the oxidation number is 1.
The oxidation number of Mn decreases from +7 to +2. The decrease in the oxidation number is +5.
Balance increase in oxidation number with decrease in oxidation number.
$H^{+} + M n O_{4}^{-} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + H_{2} O$
Balance O atoms
$H^{+} + M n O_{4}^{-} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
Balance H atoms
$8 H^{+} + M n O_{4}^{-} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
This is balanced chemical equation.
The net ionic charge on the right side of the equation $= + 2 + 5 (+ 3) = + 17$

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M n O^{4 -} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O

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