Balance(Explain clearly oxidation state of Fe3O4)

Fe3O4^- + MnO4^- + H2O=Fe2O3 + MnO2 + OH-

Asked by Vidushi412 6th October 2018, 5:48 PM

Answered by Expert

Answer:

Step 1

The skeleton equation is:

Fe₃O₄ + MnO₄^– + H₂O=Fe₂O₃ + MnO₂ + OH^–

Step 2

(a) The oxidation number of various atoms involved in the reaction.

$stack Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 space plus space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus space plus space straight H with plus 1 on top subscript 2 straight O with negative 2 on top space space rightwards arrow with equals on top space stack Fe subscript 2 with plus 3 on top straight O with negative 2 on top subscript 3 space plus space Mn with plus 4 on top stack straight O subscript 2 with negative 2 on top space plus space straight O with negative 2 on top straight H with plus 1 on top to the power of – Fe subscript 3 straight O subscript 4 space equals space Fe with plus 2 on top straight O with negative 2 on top space plus space Fe with plus 3 on top subscript 2 straight O with plus 2 on top 3 Fe space plus space 4 straight O space equals space 0 3 space straight x space plus space 4 left parenthesis negative 2 right parenthesis space equals space 0 3 straight x space equals space 8 straight x space equals space 8 over 3 space therefore space straight O. straight S. space Of space Fe space in space Fe subscript 3 straight O subscript 4 space is space 8 over 3$

(b) Identify and write out all the redox couple in reaction.

$bold Oxidation bold colon space space space stack Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 equals space Fe with plus 2 on top straight O space plus space stack Fe subscript 2 with plus 3 on top straight O subscript 3 Fe to the power of 2 plus end exponent rightwards arrow with blank on top space Fe to the power of plus 3 end exponent space plus space e to the power of minus space bold Reduction bold colon space space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus space plus space 3 space e to the power of minus rightwards arrow with blank on top space space Mn with plus 4 on top stack straight O subscript 2 with negative 2 on top Mn to the power of plus 7 end exponent space plus space 3 space e to the power of negative space end exponent rightwards arrow with blank on top space Mn to the power of plus 4 end exponent$

Step 3

Balance the atoms in each half reaction

Step 4: Make electron gain equivalent to electron lost.

$bold Oxidation bold colon space space space stack 2 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 rightwards arrow with blank on top space 3 space stack Fe subscript 2 with plus 3 on top straight O subscript 3 plus space 2 straight e to the power of minus space space space....................................... space..... space Multiply space by space 3 bold Reduction bold colon space space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus plus space 3 straight e to the power of negative space end exponent rightwards arrow with blank on top space space Mn with plus 4 on top stack straight O subscript 2 with negative 2 on top space....................................... space..... space Multiply space by space 2 bold Oxidation bold colon space space space stack 6 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 rightwards arrow with blank on top space 9 space stack Fe subscript 2 with plus 3 on top straight O subscript 3 plus space 6 space straight e to the power of minus space space space bold Reduction bold colon space space stack 2 space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus plus space 6 space straight e to the power of negative space end exponent rightwards arrow with blank on top space 2 space Mn with plus 4 on top stack straight O subscript 2 with negative 2 on top$

Step 5: Add the half-reactions together.

$stack 2 space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus plus space 6 space straight e to the power of negative space end exponent plus space stack 6 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 space rightwards arrow with blank on top space space 2 space Mn with plus 4 on top stack straight O subscript 2 with negative 2 on top space plus space space 9 space stack Fe subscript 2 with plus 3 on top straight O subscript 3 plus space 6 space straight e to the power of minus space Cancelling space electrons comma stack 2 space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus space plus space space stack 6 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 space rightwards arrow with blank on top space space 2 space Mn with plus 4 on top stack straight O subscript 2 space end subscript with negative 2 on top space space plus space space 9 space stack Fe subscript 2 with plus 3 on top straight O subscript 3$

Step 6: Simplify the equation

$A d d space r e m a i n i n g space s p e c i e s comma stack 2 space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus space plus space space stack 6 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 space space plus space H subscript 2 O space rightwards arrow with blank on top space space 2 space Mn with plus 4 on top stack straight O subscript 2 space end subscript with negative 2 on top space space plus space space 9 space stack Fe subscript 2 with plus 3 on top straight O subscript 3 space plus space OH to the power of minus Balance space hydrogen comma stack 2 space Mn with plus 7 on top straight O with negative 2 on top subscript 4 to the power of minus space plus space space stack 6 space Fe subscript 3 with 8 over 3 on top straight O with negative 2 on top subscript 4 space space plus space straight H subscript 2 straight O space rightwards arrow with blank on top space space 2 space Mn with plus 4 on top stack straight O subscript 2 space end subscript with negative 2 on top space space plus space space 9 space stack Fe subscript 2 with plus 3 on top straight O subscript 3 space plus space 2 space OH to the power of minus S i m p l i f y space t h e space e q u a t i o n space w e space g e t comma stack 2 space Mn with blank on top straight O with blank on top subscript 4 to the power of minus space plus space space stack 6 space Fe subscript 3 with blank on top straight O with blank on top subscript 4 space space plus space straight H subscript 2 straight O space rightwards arrow with blank on top space space 2 space Mn with blank on top stack straight O subscript 2 space end subscript with blank on top space space plus space space 9 space stack Fe subscript 2 with blank on top straight O subscript 3 space plus space 2 space OH to the power of minus$

Answered by Expert 8th October 2018, 1:22 PM

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