Question #6cd21

#4"HCl"(aq) + "MnO"_2(s) -> "MnCl"_2(aq) + 2"H"_2"O"(l) + "Cl"_2(g)#

So you mean #"MnO"_2(s)#? Well, that can happen only in acidic conditions, so the #"HCl"# must be dissolved already. This is generally an intermediate reaction in a permanganate test for chloride ion.

Write the half-reaction skeletons.

#"MnO"_2(s) -> "Mn"^(2+)(aq)#

#"Cl"^(-)(aq) -> "Cl"_2(g)#

• You explicitly stated the oxidation state of #"Mn"# in both compounds, so it should be obvious that manganese got reduced.

• Chlorine gas is a homonuclear diatomic, and therefore has an oxidation state that changed from #-1# in #"Cl"^(-)# to zero in #"Cl"_2#.

Now, to balance them,

1. Balance non-#"O"# and non #"H"# atoms.
2. Add #"H"_2"O"# to balance #"O"# atoms, since we are in aqueous conditions.
3. Add #"H"^(+)# to balance #"H"# atoms, since we are in acidic conditions.
4. Add #e^(-)# to balance charge, as a formal way to depict change in oxidation state.

The reduction half-reaction goes as follows. Add water.

#"MnO"_2(s) -> "Mn"^(2+)(aq) + 2"H"_2"O"(l)#

Add #"H"^(+)#.

#4"H"^(+)(aq) + "MnO"_2(s) -> "Mn"^(2+)(aq) + 2"H"_2"O"(l)#

Balance charge.

#2e^(-) + 4"H"^(+)(aq) + "MnO"_2(s) -> "Mn"^(2+)(aq) + 2"H"_2"O"(l)#

We are done with this half-reaction. Now for the oxidation half-reaction. Balance the #"Cl"#.

#2"Cl"^(-)(aq) -> "Cl"_2(g)#

Balance the charge.

#2"Cl"^(-)(aq) -> "Cl"_2(g) + 2e^(-)#

Nothing else is necessary. So, add the two half-reactions together.

#cancel(2e^(-)) + 4"H"^(+)(aq) + "MnO"_2(s) -> "Mn"^(2+)(aq) + 2"H"_2"O"(l)#
#2"Cl"^(-) -> "Cl"_2(g) + cancel(2e^(-))#
#"----------------------------------------------------------------------"#

#color(blue)(2"Cl"^(-)(aq) + 4"H"^(+)(aq) + "MnO"_2(s) -> "Mn"^(2+)(aq) + 2"H"_2"O"(l) + "Cl"_2(g))#

Or... by adding #"Cl"^(-)# on each side and recombining ions...

#4"HCl"(aq) + "MnO"_2(s) -> "MnCl"_2(aq) + 2"H"_2"O"(l) + "Cl"_2(g)#