Balance the following equation for the reaction in an acidic medium by the ion-electron method.?

Question

____Fe^2+ + ____MnO4^- ___ ______ ----->

___Fe^3+ + ___Mn^2+ + ___ ______

Answer 1

Lancenigo di Villorba (TV), Italy

I distinguish two Electronic Half-Reactions, ANODIC HALF-REACTION related to Chemical Stuffs losing ELECTRONs while CATHODIC ONE is related to Stuffs gaining ELECTRONs. I will execute a STEP-BY-STEP BALANCING upon the HALF-REACTIONs, hence I sum them in order to erase the ELECTRONs EXCHANGED.

FIRST REACTION
Can do FERROUS and PERMANGANATE IONs react one against the other one? Yes, they do.
ANODE
Fe++(aq) ---> Fe+++(aq)
this is the Anodic Half-Reaction which I will balance : I add ELECTRONs in order to balance the Oxidation Number in the Chemical Elements able to vary them
Fe++(aq) ---> Fe+++(aq) + e
and this is balanced one.
CATHODE
MnO4-(aq) ---> Mn++(aq)
this is the Cathodic Half-Reaction which I will balance : I add ELECTRONs in order to balance the Oxidation Number in the Chemical Elements able to vary them
MnO4-(aq) + 5e ---> Mn++(aq)
I add WATER in order to balance the Oxygen Atoms
MnO4-(aq) + 5e ---> Mn++(aq) + 4 H2O(aq)
I add HYDROGEN IONs in order to balance the Hydrogen Atoms
MnO4-(aq) + 8 H+(aq) + 5e ---> Mn++(aq) + 4 H2O(aq)
and this is balanced one.
REACTION's SUM
Anodic Half-Reaction exchanges ONE ELECTRON while Cathodic One exchanges FIVE ONEs.
Thus, I combine them by multiplying in order to erase the EXCHANGED ELECTRONs
(X5) Fe++(aq) ---> Fe+++(aq) + e
(X1) MnO4-(aq) + 8 H+(aq) + 5e ---> Mn++(aq) + 4 H2O(aq)
retrieving the following one
5 Fe++(aq) + MnO4-(aq) + 8 H+(aq) --->
---> 5 Fe+++(aq) + Mn++(aq) + 4 H2O(aq)

SECOND REACTION
Can do FERRIC and MANGANOUS IONs react one against the other one? Yes, they do.
CATHODE
Fe+++(aq) ---> Fe++(aq)
this is the Cathodic Half-Reaction which I will balance : I add ELECTRONs in order to balance the Oxidation Number in the Chemical Elements able to vary them
Fe+++(aq) + e ---> Fe++(aq)
and this is balanced one.
ANODE
Mn++(aq) ---> MnOOH(s)
this is the Anodic Half-Reaction which I will balance : I add ELECTRONs in order to balance the Oxidation Number in the Chemical Elements able to vary them
Mn++(aq) ---> MnOOH(s) + e
I add WATER in order to balance the Oxygen Atoms
Mn++(aq) + 2 H2O(aq) ---> MnOOH(s) + e
I add HYDROGEN IONs in order to balance the Hydrogen Atoms
Mn++(aq) + 2 H2O(aq) ---> MnOOH(s) + 3 H+(aq) + e
and this is balanced one.
REACTION's SUM
Anodic Half-Reaction exchanges ONE ELECTRON as Cathodic One does.
Thus, I combine them by multiplying in order to erase the EXCHANGED ELECTRONs
(X1) Fe+++(aq) + e ---> Fe++(aq)
(X1) Mn++(aq) + 2 H2O(aq) ---> MnOOH(s) + 3 H+(aq) + e
retrieving the following one
Fe+++(aq) + Mn++(aq) + 2 H2O(aq) --->
---> Fe++(aq) + MnOOH(s) + 3 H+(aq)

I hope this helps you.

Answer 2

Answering question 2 first, OH- is the component of water that shows you're in a difficulty-unfastened answer, and not an acid answer. If the region have been reversed, you will see H+. The reaction is Cr(OH)3 + H2O2 -> CrO4= + OH- First, you become responsive to the species being oxidized and the species being decreased. the single being oxidized is Cr+3, which suits to Cr+6 interior the chromate ion. the single being decreased is the O-a million (peroxide is H-O-O-H), which suits to O=. stability the oxidation and help reactions. grant H2O as mandatory to stability the O and H standards. Cr(OH)3 + 5OH- -> CrO4 = + 4 H2O +3e H2O2 + 2e -> 2 OH- We multiply the Cr(OH)3 eqtn with the help of two and the H2O2 equation with the help of three to get 2 Cr(OH)3 +10 OH- -> 2 CrO4= + 8 H2O + 6e 3 H2O2 + 6e -> 6 OH- We upload equations and pass basic species to the realm the place they're constructive-valued. 2 Cr(OH)3 + 4OH- + 3 H2O2 -> 2 CrO4= + 8H2O

Answer 3

(MnO4)- + 8H+ + 5e- ------> Mn2+ + 4H2O

Fe2+ -----> Fe3+ + e-

Now multiply the whole of the second equation by 5, and add it to the first one, missing out the electrons.