balance the following equation by ion electron method in acidic medium [MnO4]- + [Fe]2+ = [Mn]2+ [Fe]3+

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Solution

$M n {O_{4}}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +} S p l i t i n t o h a l f r e a c t i o n s M n {O_{4}}^{-} \to M n^{2 +} F i n d t h e o x i d a t i o n s t a t e o f M n i n b o t h t h e s i d e s M n {O_{4}}^{-} x + 4 (- 2) = - 1 x = + 7 M n^{2 +} x = + 2 T h u s 5 e l e c t r o n s a r e g a i n e d M n {O_{4}}^{-} + 5 e^{-} \to M n^{2 +} B a l a n c e t h e o x y g e n b y i n t r o d u c i n g H_{2} O M n {O_{4}}^{-} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O B a l a n c e t h e h y d r o g e n M n {O_{4}}^{-} + 5 e^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O - - - - 1 C o n s i d e r t h e s e c o n d h a l f r e a c t i o n F e^{2 +} \to F e^{3 +} F e^{2 +} \to F e^{3 +} + 1 e^{-} - - - - - 2 N o w a d d 1 a n d 2 . M n {O_{4}}^{-} + 5 e^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O F e^{2 +} \to F e^{3 +} + 1 e^{-} M u l t i p l y t h e s e c o n d e q u a t i o n b y 5 a s e l e c t r o n s a r e n o t b a l a n c e d . M n {O_{4}}^{-} + 5 e^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O 5 F e^{2 +} \to 5 F e^{3 +} + 5 e - - - - - - - - - - - - - - - - - M n {O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O + 5 F e^{3 +} - - - - - - - - - - - - - - - - - - - - - - - - - - - T h u s t h e e q u a t i o n i s a b a l a n c e d e q u a t i o n .$

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