Chemistry Reference
In-Depth Information
Solution
2
Sn
2
(aq) £ Cr
3
(aq)
Cr
2
O
(aq)
SnO
2
(s)
?
7
The reduction half-reaction is presented first:
Step 1:
2
(aq) £ Cr
3
(aq)
Cr
2
O
7
2
(aq) £ 2 Cr
3
(aq)
Cr
2
O
7
(The chromium is balanced first.)
Step 2:
No change.
2
(aq) £ 2 Cr
3
(aq)
7 H
2
O(
/
)
Step 3:
Step 4:
Step 5:
Cr
2
O
7
14 H
(aq)
2
(aq) £ 2 Cr
3
(aq)
Cr
2
O
7 H
2
O(
/
)
7
6 e
14 H
(aq)
2
(aq) £ 2 Cr
3
(aq)
Cr
2
O
7 H
2
O(
/
)
7
(Balanced half-reaction)
Step 6:
(a) Net charge on each side
(b) 14 H, 7 O, 2 Cr on each side
(c) Oxidation number change for
two atoms
from
6
6
to
3
is
equal to the number of electrons.
The oxidation half-reaction is then worked out:
Step 1:
Step 2:
Sn
2
(aq) £ SnO
2
(s)
No change.
Sn
2
(aq) £ SnO
2
(s)
Step 3:
2 H
2
O(
/
)
Sn
2
(aq) £ SnO
2
(s)
4 H
(aq)
2 H
2
O(
/
)
Step 4:
Step 5:
Sn
2
(aq) £ SnO
2
(s)
4 H
(aq)
2 e
2 H
2
O(
/
)
(Balanced half-reaction)
Step 6:
Everything checks.
Step 7:
Make the number of electrons in each the same by multiplying the
oxidation half-reaction by 3:
3 Sn
2
(aq) £ 3 SnO
2
(s)
12 H
(aq)
6 e
6 H
2
O(
/
)
Combine the two half-reactions,
3 Sn
2
(aq)
14 H
(aq)
2
6 H
2
O(
/
)
Cr
2
O
(aq) £
7
12 H
(aq)
2 Cr
3
(aq)
3 SnO
2
(s)
7 H
2
O(
/
)
12 H
Step 8:
Eliminating
and
6 H
2
O
from both sides yields the complete
balanced equation:
3 Sn
2
(aq)
2 H
(aq)
2
Cr
2
O
(aq) £
7
2 Cr
3
(aq)
H
2
O(
/
)
3 SnO
2
(s)
Step 9:
6
net charge on each side, 3 Sn, 2 H, 2 Cr, 7 O
Practice Problem 16.13
Complete and balance the equation for the
reaction of the permanganate ion with chloride ion in acid solution to give man-
ganese(II) ion and perchlorate ion, as well as other products:
MnO
4
Cl
(aq) £ Mn
2
(aq)
ClO
4
(aq)
(aq)
?
