Question

Refer to the given oxidation-reduction reaction, which takes place in acidic solution:
MnO₄⁻(aq) + Br⁻(aq) → MnO₂(s) + BrO₃⁻(aq)
The number of electrons transferred in the reaction is: 9?

Answer 1

no

Answer 2

help me out lol

Answer 3

you need only the answer or you have to know how to do it?

Answer 4

both

Answer 5

1 strat for calculating the oxidation states of the elements in the equation and figuring out what element oxidize and which element reduce

Answer 6

Mn is going from a +7 to a +4 and Br is going from a -1 to a +5?

Answer 7

MnO₄⁻(aq) , O is -2, then -2x4 =-8, -8-(-1) =-7, then Mn is +7
Br⁻(aq) it is an anion with one negative charge, the oxidation state is -1
MnO₂(s) O is -2, then -2x2 =-4, then Mn here is +4
calculate BrO₃⁻(aq)

Answer 8

So whats next

Answer 9

calculate BrO₃⁻(aq)

Answer 10

I already did both

Answer 11

which element get oxidized and witch one get reduced?

Answer 12

write down the two half reaction and balance them

Answer 13

Br gets oxidized and Mn gets reduced

Answer 14

Mn is going from a +7 to a +4 and Br is going from a -1 to a +5

Answer 15

Good! now writeh down the two equation and balance the half equations
Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:

Balance elements in the equation other than O and H.
Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.
Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.
Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.)
The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.
(If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.)
The equation can now be checked to make sure that it is balanced.
http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions

Answer 16

You might have lost me with all this but i will try

Answer 17

\[3e ^{-}+4H ^{+}+MnO _{4}^{-} \rightarrow MnO _{2}+2H _{2}O\]

Answer 18

\[3H _{2}O+Br ^{-} \rightarrow BrO _{3}^{-}+6H ^{+}+6e ^{-}\]

Answer 19

the first half reaction exchange 3e-, the second one 6e-, then you will need the double of the first reaction to react with the second reaction, Multiply the first reaction by 2 and add the two reactions to get the balanced equation. Here you see, that when the reaction is balanced 6e- and not 9e has you said at the beginning are transferred.

Answer 20

Dude you dont understand how much I appreciate you helping me out. You just went through all that trouble to help me understand, i would give you 100 medals if i could.

Answer 21

did you got it? it is not the most easy topic

Answer 22

I actually knew how to do this but forgot, but thanks to you working this out step by step in combination with the detailed notes i had i now understand how to do it again

Answer 23

best wishes!!