$M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +} + H_{2} O$ . (acidic medium)
Balanced reaction is:
$M n O_{4}^{-} + a F e^{2 +} + b H^{+} \to M n^{2 +} + c F e^{3 +} + d H_{2} O$ .

a+b+c+d is:

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Solution

$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ .

$a = 5$
$b = 8$
$c = 5$
$d = 4$
$a + b + c + d = 22$

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Similar questions

F e^{2 +} \to F e^{3 +} + e^{-}; M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}

, the ratio of stoichiometric coefficient of

F e^{2 +}

and

M n O_{4}^{-}

in the balanced redox reaction is:

Balance the following equation:

{M n O}_{4}^{-} \to M n^{2 +} + H_{2} O

Q. For the following redox reaction:

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O

The correct stoichiometric coefficients of

M n O_{4}^{-}, C_{2} O_{4}^{2 -} a n d H^{+}

are, respectively:

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

Q. While

F e^{3 +}

is stable,

M n^{3 +}

is not stabe in acid solution because:

(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)

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MnO−4+Fe2+→Mn2++Fe3++H2O. (acidic medium)Balanced reaction is:MnO−4+aFe2++bH+→Mn2++cFe3++dH2O.a+b+c+d is:

MnO−4+5Fe2++8H+→Mn2++5Fe3++4H2O.a=5b=8c=5d=4 a+b+c+d=22

$M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +} + H_{2} O$ . (acidic medium)
Balanced reaction is:
$M n O_{4}^{-} + a F e^{2 +} + b H^{+} \to M n^{2 +} + c F e^{3 +} + d H_{2} O$ .

a+b+c+d is:

$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ .

$a = 5$
$b = 8$
$c = 5$
$d = 4$
$a + b + c + d = 22$