In: Chemistry
Balance the following oxidation reduction reaction under acidic conditions. Show your work. MnO4− + H2SO3→ SO4 2− + Mn2+
reduction :
MnO4- ----------------------------> Mn+2
(+7) (+2)
total 5 electrons are transfered
MnO4- +5e- ----------------------------> Mn+2
balance oxygens :
MnO4- + 8 H+ + 5e- ----------------------------> Mn+2 + 4H2O (equation 1)
oxidation :
H2SO3 -------------------------> SO4^-2
(+4) (+6)
two electrons are transfered
H2SO3 + 2e- -------------------------> SO4^-2
balance oxygens
H2SO3 + H2O + 2e- -------------------------> SO4^-2 + 4H+ (equation 2)
multiply equation (1) with 2 and multiply equation (2) with 5 and than add
2MnO4- + 16 H+ + 5H2SO3 + 5H2O ------------------> 2Mn+2 + 8H2O + 20 H+ + 5 SO4^-2
2MnO4- + 5H2SO3 + ------------------> 2Mn+2 + 3H2O + 4 H+ + 5 SO4^-2