Question

b. Assume a reaction takes place in a basic solution to form the given products:
MnO4–(aq) + Cl–(aq) MnO2(s) + Cl2(g) (unbalanced)
      i. Balance the given half-reactions for atoms and charge. (.5 point)
MnO4– + H2O MnO2 + OH–
Cl– Cl2
      ii. Multiply to balance the charges in the reaction. (.5 point)
      iii. Add the equations and simplify to get a balanced equation. (.5 point)

Answer 1

so here's what i think:
1. half-reactions:
MnO2 --> MnO4-
Cl2 --> Cl-
2. 2H2O + MnO2 --> MnO4- + 4H+ 3e-
2e- + Cl2 --> 2Cl-
3.8OH- + 2MnO2 + 3Cl2 --> 2MnO4- + 6Cl- + 4H2O

Answer 2

is this correct? thanks!

Answer 3

@JFraser

Answer 4

it will take me a few minutes to work it through myself, but it looks ok at first

Answer 5

ok

Answer 6

wait no ii would be: 4H2O + 2MnO2 --> 2MnO4- + 8H+ + 6e-
6e- + 3Cl2 --> 6Cl-

Answer 7

no wait i messed this up hold on

Answer 8

1. 2H2O + MnO2 --> MnO4- + 4H+ 3e-
2e- + Cl2 --> 2Cl-
2. 4H2O + 2MnO2 --> 2MnO4- + 8H+ + 6e-
6e- + 3Cl2 --> 6Cl-
3. 8OH- + 2MnO2 + 3Cl2 --> 2MnO4- + 6Cl- + 4H2O

Answer 9

since the half reactions in number one have to be balanced

Answer 10

this set of half-reactions doesn't fit together. the electrons won't cancel since they're both on the same side

Answer 11

oh. i must've messed up.

Answer 12

i'll start with the chlorine, since that's easier

Answer 13

here's the exact question:
b. Assume a reaction takes place in a basic solution to form the given products:
MnO4–(aq) + Cl–(aq) MnO2(s) + Cl2(g) (unbalanced)
      i. Balance the given half-reactions for atoms and charge. (.5 point)
      ii. Multiply to balance the charges in the reaction. (.5 point)
      iii. Add the equations and simplify to get a balanced equation. (.5 point)

Answer 14

you start with \[Cl_2 \rightarrow Cl^{-1}\]balance the Cl's first:\[Cl_2 \rightarrow 2Cl^{-1}\]balance O's and H's, but there are none, so we skip to balancing charge\[2e^{-1} + Cl_2 \rightarrow Cl^{-1}\]

Answer 15

ok, now that's different

Answer 16

the first reaction had Cl2 --> Cl-

now it's written Cl- --> Cl2

Answer 17

i thought it's 2e- + Cl2 --> 2Cl-

Answer 18

the REDUCTION is written with the gain of electrons, but you have to use the reaction the direction the reaction tells you

Answer 19

i must've typed something wrong hold on

Answer 20

use the half-reaction the way it's given, don't flip it backwards

Answer 21

ah i went backward

Answer 22

i switched the reduction

Answer 23

yea okay so i need to fix that

Answer 24

\[Cl_2 \rightarrow 2Cl^{-1} + 2e^{-1}\]

Answer 25

the other half-reaction is more complicated, so do it one step, by the instructions

\[MnO_2 \rightarrow MnO4^{-1}\]

Answer 26

balance atoms other than O and H: already done, so next step

Answer 27

balance O atoms with water:

\[2H_2O + MnO_2 \rightarrow MnO_4^{-1}\]

Answer 28

balance H atoms with H+

\[2H_2O + MnO_2 \rightarrow MnO_4^{-1} + 4H^{+1}\]

Answer 29

hold on i think i understand

Answer 30

if the solution were acidic, we'd be almost done, but the problem says the solution is BASIC

Answer 31

this is iii 2MnO4^- + 4H2O + 6Cl- --> 2MnO2 + 8OH- + 3Cl2

Answer 32

i got it!!! i think i understand it

Answer 33

i'll type out i and ii

Answer 34

H+ ions don't float around in basic solutions, so we get rid of them by mixing in equal numbers of OH- to make water

\[4OH^{-1} + 2H_2O + MnO_2 \rightarrow MnO_4^{-1} + (4H^{+1} + 4OH^{-1})\]

Answer 35

so for i it would be MnO4^- + 4H+ --> MnO2 + 2H2O but MnO4^- + 2H2O --> MnO2 + 4OH- is the answer because it needs to be basic

Answer 36

don't flip it around, use the half-reaction in the direction the problem tells you. See how you've flipped yours backwards from mine?

Answer 37

wait, hold on i did it again huh

Answer 38

yup, you did

Answer 39

but i did that so it's for the acidic one and then you can fix it in the end so it becomes basic.

Answer 40

look at the last one i wrote. adding hydroxides is how we fix a half-reaction so that it's basic

Answer 41

i think I've got it from here. thanks

Answer 42

let me know what you come up with, i have my answer ready

Answer 43

the problem was that the half-reactions were already given

Answer 44

i completely forgot to copy that

Answer 45

the half-reactions will always be given, you have to put them together

Answer 46

i mean my teacher solved i for us so anyways i understand

Answer 47

\(4H_2O + 2MnO_2 + 10Cl^{-1} \rightarrow 2MnO_4^{-1} + 8OH^{-1} + 5Cl_2\)