Sarah asked in Science & MathematicsChemistry · 1 month ago

# Identify the oxidation state, write the half reaction, balance the reaction, and identify the reducing and oxidizing agent ?

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• Dr W
Lv 7
1 month ago

I cover balancing redox reactions in my answer here.

remember 2 things from this

.. (1) reduction = reduction in oxidation state

.. (2) agents are opposites

anyway

*** oxidation states ***

on the left

.. Mn in MnO4- is +7.... why? O is -2, the overall is -1.. so 1x(Mn)+4*(-2) = -1

.. O in MnO4- is -2.. why?  O has the higher electronegativity and gets 2 electrons

.. S in SO3(-2) is +4.. why?  O is -2, overall is -2.. 1xS + 3*(-2) = -2 --> S=+4

.. O in SO3(-2) is -2.... O has the higher EN and gets the 2 electrons to fill it's orbital

on the right

.. Mn in MnO2(s) is +4.. O is -2, overall is 0.. 1*Mn + 2(-2) = 0 --> Mn = +4

.. O in MnO2 is -2... O with the higher EN gets all the electrons it needs.

.. S in SO4(-2)) is +6.. O is -2, 1*S+4(-2) = -2... S = +6

.. O in SO4(-2) is -2.. O gets the 2 electrons to fill its outer orbital

*** half reactions ***

Mn(7+) + 3 e's ---> Mn(4+).. .. . .. reduction half.. (ox. state is reduced right?)

S(4+) ----> S(6+) + 2 e's... . .. . . ..oxidation half

*** balancing ***

step (1) identify oxidation state.. done

step (2) determine what is oxidized and what is reduced.. .done

step (3) write half reactions.... .done

step (4) balance e's

.. 2 Mn(7+) + 6 e's ---> 2 Mn(4+)

.. 3 S(4+) ---> 3 S(6+) + 6 e's

step (5) add half reactions and cancel e's

.. 2 Mn(7+) + 3 S(4+) + 6 e's ---> 2 Mn(4+) + 3 S(6+) + 6 e's

.. 2 Mn(7+) + 3 S(4+) ---> 2 Mn(4+) + 3 S(6+)

step (6) rearrange and add counter ions as necessary

.. 2 MnO4(-) + 3 SO3(2-) ---> 2 MnO2(s) + 3 SO4(2-)

*** identify oxidizing and reducing agent

Mn went from +7 to +4 was reduced and is the oxidizing agent

S went from +4 to +6 was oxidized and is the reducing agent