Question

with both half - reactions now balanced and having the same number of electrons transferred, they can be combined.
$6e^- + 4h_2o + 2mno_4^- \
ightarrow 2mno_2 + 8oh^-$
$6oh^- + br^- \
ightarrow bro_3^- + 3h_2o + 6e^-$
how many $oh^-$ ions are on the reactant side in the combined, overall equation?
$h_2o + 2mno_4^- + br^- \
ightarrow 2mno_2 + bro_3^- + ?oh^-$

Explanation:

\times2=6\) H (from \(3H_2O\)). So H is balanced in second half - reaction.

Now, when we add the two half - reactions:

Left side H: 8 (from first) + 6 (from second) = 14

Right side H: 8 (from first) + 6 (from second) = 14. So H is balanced.

Now, let's write the combined reaction:

\(6e^-+4H_2O + 2MnO_4^-+6OH^-+Br^-
ightarrow2MnO_2 + 8OH^-+BrO_3^-+3H_2O + 6e^-\)

Cancel \(6e^-\):

\(4

Answer:

\times2=6\) H (from \(3H_2O\)). So H is balanced in second half - reaction.

Now, when we add the two half - reactions:

Left side H: 8 (from first) + 6 (from second) = 14

Right side H: 8 (from first) + 6 (from second) = 14. So H is balanced.

Now, let's write the combined reaction:

\(6e^-+4H_2O + 2MnO_4^-+6OH^-+Br^-
ightarrow2MnO_2 + 8OH^-+BrO_3^-+3H_2O + 6e^-\)

Cancel \(6e^-\):

\(4