Using the method of half reactions to balance a redox equation
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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