Using the method of half reactions to balance a redox equation - Oakton

Using the method of half reactions to balance a redox equation

Balance: Cl-(aq) + Cr2O72-(aq) Cl2(g) + Cr3+ (in acidic solution).

(NOTE: To save space and avoid "clutter", I will leave off state designations until the end) gains e-`s

Cr2O72- + Cl- Cr3+ + Cl2

+6 -2

-1

+3

0

(acidic)

loses e-`s

Reduction Half Reaction (I show each step so you can see the order of doing things, but

if I were doing this on an exam, I would not show each step separately!)

Cr2O72- 2 Cr3+

(balance all non O's and H's)

Cr2O72- 2 Cr3+ + 7 H2O (balance the O's with H2O)

Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O (balance the H's with H+'s)

+12

+6 => need 6 e-`s on left

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Helpful Hint: Always add electrons to the more positive side to bring the charge

"down to" the other side's

(since electrons are negative! )

Oxidation Half Reaction

2 Cl- Cl2 (balance non O's and H's first; no O's or H's to balance!)

-2

0 => need 2 e-`s on right

2 Cl- Cl2 + 2 e-

Overall Equation

To make the number of electrons transferred equal, just take the oxidation half reaction and

x 3

multiply it by THREE (to get 6 e-`s total produced--same as the number needed in the

reduction half reaction). Note that this is an "easy" one (that is, you don't need to multiply

BOTH equations by a factor to make the number of electrons equal); make sure that you are

confident that you could do one like we did in class where the least common multiple is neither

of the values of electrons already "given".

6 Cl- 3 Cl2 + 6 e-

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Cr2O72- + 14 H+ + 6 Cl- 3 Cl2 + 2 Cr3+ + 7 H2O

Just to show you how it would look if you were to "convert" the H+'s into H3O+'s like the text chooses to do, at this point you would have to add 14 H2O's to both sides to yield:

Cr2O72- + 14 H3O+ + 6 Cl- 3 Cl2 + 2 Cr3+ + 21 H2O

Adding state designations: Cr2O72-(aq) + 14 H+(aq) + 6 Cl-(aq) 3 Cl2(g) + 2 Cr3+(aq) + 7 H2O(l)

OR Cr2O72-(aq) + 14 H3O+(aq) + 6 Cl-(aq) 3 Cl2(g) + 2 Cr3+(aq) + 21 H2O(l)

Check charge on both sides: (-2) + (+14) + (-6) = +6 (left) and 2(+3) = +6 (right)

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