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Electrochemistry II Balancing Equations
Often the oxidized form of an atom is combined with oxygen, whereas in the
reduced state it is combined with less oxygen, or none Chromium is a typical
example, in the Cr
8+ state it is combined as Cr 2 Of~, but in the Cr
3
"
1
" state it is
uncombmed (except for hydration) In working out a suitable half-reaction
equation, you must decide what to do with the oxygen atoms The answer is
simple you may use H
+ , OH~, and H 2 O on either side of the equations for
balancing, so long as you comply with the actual state of acidity of the solutions. If a solution is acidic, you must not use OH~ on either side of an equation
for balancing, you must use H
+ and/or H 2 O.
PROBLEM:
Write a balanced half-reaction for the oxidation of metallic gold to its highest
oxidation state
SOLUTION:
Like all elements in the uncombmed state, metallic gold (Au) has a charge of 0
There is no simple way in which you can reason out the fact that gold s highest
oxidation state is +3, presumably you learned this in Chapter 8 Now, knowing
the two oxidation states involved, you can write your half-reaction as
3e~ + Au
3+ *± Au
As a check we note that the sum of the electrical charges on each side is zero
PROBLEM:
Write a balanced half-reaction equation for the oxidation of Mo
3+ to MoO^ in acid
solution
SOLUTION:
You are not expected to know about the chemistry of Mo but, once you are given
the reactant and product, there is no difficulty For the MoO^ ion,
C = +1 = (l)(z Mo ) + (2)(z 0 )
= Z MO + (2)(-2)
Z Mo = +5
Because the oxidized form contains oxygen and the reduced form does not, and
because the solution is acidic, the most direct approach here is to add sufficient H
+
to combine with the oxygen atoms to form water, in this case 4H
+ would combine
with 2 O
2 ~ to make 2 H 2 O Note that the H
+ and O
2 ~ do not involve any changes in
charge We write
2e- + 4H
+ + MoO 2
+ f± 2H 2 O + Mo
3+
As a check, we note that the sum of the electncal charges on each side is +3
Electrochemistry II Balancing Equations
Often the oxidized form of an atom is combined with oxygen, whereas in the
reduced state it is combined with less oxygen, or none Chromium is a typical
example, in the Cr
8+ state it is combined as Cr 2 Of~, but in the Cr
3
"
1
" state it is
uncombmed (except for hydration) In working out a suitable half-reaction
equation, you must decide what to do with the oxygen atoms The answer is
simple you may use H
+ , OH~, and H 2 O on either side of the equations for
balancing, so long as you comply with the actual state of acidity of the solutions. If a solution is acidic, you must not use OH~ on either side of an equation
for balancing, you must use H
+ and/or H 2 O.
PROBLEM:
Write a balanced half-reaction for the oxidation of metallic gold to its highest
oxidation state
SOLUTION:
Like all elements in the uncombmed state, metallic gold (Au) has a charge of 0
There is no simple way in which you can reason out the fact that gold s highest
oxidation state is +3, presumably you learned this in Chapter 8 Now, knowing
the two oxidation states involved, you can write your half-reaction as
3e~ + Au
3+ *± Au
As a check we note that the sum of the electrical charges on each side is zero
PROBLEM:
Write a balanced half-reaction equation for the oxidation of Mo
3+ to MoO^ in acid
solution
SOLUTION:
You are not expected to know about the chemistry of Mo but, once you are given
the reactant and product, there is no difficulty For the MoO^ ion,
C = +1 = (l)(z Mo ) + (2)(z 0 )
= Z MO + (2)(-2)
Z Mo = +5
Because the oxidized form contains oxygen and the reduced form does not, and
because the solution is acidic, the most direct approach here is to add sufficient H
+
to combine with the oxygen atoms to form water, in this case 4H
+ would combine
with 2 O
2 ~ to make 2 H 2 O Note that the H
+ and O
2 ~ do not involve any changes in
charge We write
2e- + 4H
+ + MoO 2
+ f± 2H 2 O + Mo
3+
As a check, we note that the sum of the electncal charges on each side is +3