SAT Subject Test Chemistry,10 edition

Finally,    add H+  to  the left    side    to  balance the 4H 2 Os.    These   two half-reactions  are

now balanced    in  mass    (but    not in  charge).

MnO 4 − +   8H+ →   Mn2+    +   4H 2 O

Step
3:

Balance the charges of  each    half-reaction.  The reduction   half-reaction   must    consume the

same    number  of  electrons   as  are supplied    by  the oxidation   half.   For the oxidation

reaction,   add 2   electrons   to  the right   side    of  the reaction:

2I− → I 2   +   2e−

For the reduction   reaction,   a   charge  of  +2  must    exist   on  both    sides.  Add 5   electrons

to  the left    side    of  the reaction    to  accomplish  this:

5e− +   8H+ +   MnO 4 − →   Mn2+    +   4H 2 O

Step
4:

Both    half-reactions  must    have    the same    number  of  electrons   so  that    they    will    cancel.

Multiply    the oxidation   half    by  5   and the reduction   half    by  2   and add the two:

5(2I− → I 2     +   2e−)

2(5e−   +   8H+ +   MnO 4 − →   Mn2+    +   4H 2 O)

The final   equation    is:

10I−    +   10e−    +   16H+    +   2MnO 4 − →  5I 2    +   2Mn2+   +   10e−    +   8H 2 O

To  get the overall equation,   cancel  out the electrons   and any H 2 Os, H+s,    OH−s,   or  e−s

that    appear  on  both    sides   of  the equation.

10I−    +   16H+    +   2MnO 4 − →  5I 2    +   2Mn2+   +   8H 2 O