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Electrochemistry II Balancing Equations
Preliminary coefficients of 2 are used in front of Br~ and Cr
3+ because,
no matter what the final coefficients, there will always be 2Br~ for every
Br 2 and 2Cr
3+ for every Cr 2 O|~ Cr and Br are the elements involved in
electron transfer
(2) Determine the loss and gain of e~ as Crgoes from +6 to +3, and Br goes
from -1 to 0
gam = (2)(3e-) = 6e~
2Br->
2Cr
3+
+
Br s
loss = (2)(le-) = 2e~
(3) The least common multiple for 6 and 2 is 6 Provide an equal loss and
gain of e~ by multiplying Cr 2 O^~ and 2Cr
3+ by 1, and 2Br~ and Br 2
by 3, to give
gam = (2)(3e-) = 6e~
Cr 2 Of+
6Br~
-»
2Cr
3+
+
3Br 2
loss = (3)(2)(le-) = 6e~
(4) The sum of the ionic charges on the left side is -8, the sum on the
nght is +6 Addition of 14 H
+ to the left or 14 OH~ to the nght
would balance the ionic charges, but we must use 14 H
+ because the
solution is acidic This gives
Cr 2 O 7
2 ~ + 6Br- + 14 H
+ -> 2Cr
3+ + 3Br 2
(5) There are 7 O atoms on the left, and none on the nght, so we must add
7H 2 O to the right to give the balanced equation
Cr 2 O
2 - + 6Br- + 14 H
+ -> 2Cr
3+ + 3Br 2 + 7H 2 O
PROBLEMBalance the equation for the reaction between MnO^ (from KMnO 4 ) and
(from Na 3 AsO 3 ) in basic solution
SOLUTION:
We follow the stepwise procedure as before
(1) Write down the oxidizing and reducing agents, and their corresponding
reduced and oxidized forms
MnO 4 - + AsO 3
3 - -> MnO 2 + AsO 4
3
-
No preliminary coefficients are needed
(2) Determine the loss and gam of e~ as Mn goes from +7 to +4, and
As goes from +3 to +5