In order to answer this, you must first split each reaction into its half reactions:
(reduced because Br goes from +5 to -1)
(oxidized because Sn goes from 2+ to 4+)
now you would balance the coefficients of each main element, but they are already balanced (1 Br on each side and 1 Sn on each side) so now you must add
for every
present in the half reactions:
there aren't any O for Sn, so you ignore that for now. In order to balance the H you just added with the water, you must add the same amount of H added to the other side
now that all the O and H have been balanced, you figure out what the total charge on each side of the half reactions are
has a charge of
on the right and
on the left, so you must add 6 electrons to the right side to make the charges equal
for the other half reaction
has a charge of
on the right side and
on the left side, so you must add 2 electrons to the right side to make the charges equal
now you must multiply each of the half reactions by a coefficient that makes the number of electrons equal so you can cancel them
to give you
now you can add the equations and cancel those items that appear on both sides of the arrow
you can check this by counting each of the elements to make sure they are equal on both sides:
Br- 1 on left and 1 on right
O- 3 on left and 3 on right
H - 6 on left and 6 on right
Sn - 3 on left and 3 on right
I hope this helps!