Question

a. The following reaction takes place in an acidic solution.
MnO4–(aq) + Cl–(aq) Mn2+ + Cl2(g) (unbalanced)
      i. Write the reduction and oxidation half-reactions (without electrons). (.5 point)
      ii. Balance the equations for atoms (except O and H). (.5 point)
      iii. Balance the equations for atoms O and H using H2O and H+. (.5 point)
      iv. Balance the charge in the half-reactions. (.5 point)
    v. Multiply each half-reaction by the proper number to balance charges in the reaction. (.5 point)

Answer 1

Correct me if I'm wrong, thanks.
i. MnO4^-(aq) → Mn^2+
Cl^- (aq)→ Cl2(g)
iii. MnO4^- → Mn^2+ + 4H2O (balance O)
MnO4^- + 8H^+ → Mn^2+ + 4H2O (balance H)
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O

Answer 2

Looks good so far to me.

Answer 3

alright thanks. how do i do number 2... I'm not sure

Answer 4

The Mn one is already balanced since you have 1 Mn in the reactants and products.

Answer 5

The Cl one needs balancing though.2Cl^- -> Cl2

Answer 6

2Cl^- → Cl2 + 2e^-

Answer 7

wait but balanced ones can't have e...?

Answer 8

if they are balanced, then there's nothing to do with them

Answer 9

We are just to balance it for atoms for ii

Answer 10

okay

Answer 11

okay so here's what I've go so far:
i. MnO4^-(aq) → Mn^2+
Cl^- (aq)→ Cl2(g)
ii. MnO4^-(aq) → Mn^2+
2Cl^- → Cl2 + 2e^-
iii. MnO4^- → Mn^2+ + 4H2O (balance O)
MnO4^- + 8H^+ → Mn^2+ + 4H2O (balance H)
iv. MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O
2Cl^- → Cl2 + 2e^-

Answer 12

now that each half-reaction is balanced by itself, you can look at adding the two half-reactions together. The ONE THING you have to be careful of is that there are no "leftover" electrons after you've added the two half-reactions together.

that's what it's asking in part v. If you notice, the first half-reaction uses 5 electrons, but the chlorine half-reaction only uses 2. You have to get the electrons on each side to cancel before adding the half-reactions together

Answer 13

2MnO4^-(aq) + 16H^+(aq) + 10Cl^-(aq) → 2Mn^2+(aq) + 5Cl2(aq) + 8H2O(l)

Answer 14

right?

Answer 15

that's what I get, GJ

Answer 16

thanks. okay but for ii, I'm still confused. and for iii am i supposed to balance both equations...?

Answer 17

did i even do iii right?

Answer 18

you've already done the whole thing, it's solved

Answer 19

oh okay thanks

Answer 20

Looks perfect to me too :)

Answer 21

http://finedrafts.com/files/CUNY/chemistry/General/Zumdahl%20etal/8th%20ed/CH18%20Electrochemistry.pdf

galvanic cells woohooo I love this

Answer 22

thanks everyone! I really appreciate it :)

Answer 23

YW

Answer 24

No worries

Answer 25

sorry @JFraser so i handwrote this and i think i made a mistake in either ii or iii because i must've balanced Cl^- (aq)→ Cl2(g) for atoms(except O and H) wrong...

Answer 26

in the chlorine half-reaction there are no H or O atoms, so you don't have to worry about them

Answer 27

i thought its 2C^- --> Cl2 + 2e^- but i got that for iii also...

Answer 28

for the chlorine half-reaction, parts ii and iii are the same.

Answer 29

wait so sorry now I'm confused. what's the answer for ii because i know the Mn equation is already balanced(except for O and H)

Answer 30

oh

Answer 31

the steps take you sequentially through the process. if there are no O atoms, and no H atoms, steps ii and iii will be identical

Answer 32

ah okay that makes sense thanks!