How do we balance the oxidation of chloride anion to give chlorine gas by potassium permanganate?

2 Answers
May 28, 2017

Answer:

See diagram in explanation

Explanation:

Oxidation-Reduction Rxn => #MnO_2# in acidic medium.
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May 28, 2017

Answer:

We use the redox method of balancing equations, where electrons are introduced as virtual particles...........

Explanation:

Using oxidation numbers we write the reduction equation:

#stackrel(IV)MnO_2(s)+4H^+ + 2e^(-)rarrMn^(2+)+2H_2O# #(i)#

Both mass and charge are balanced as required.........

And then we need a reduction half equation.........

#Cl^(-) rarr 1/2stackrel(0)Cl_2+e^-# #(ii)#

Any species whose #"oxidation number"# #"INCREASES"# is said to be #"oxidized"#, and any species whose #"oxidation number"# #"decreases"# is said to be #"reduced"#,

We add #(i)# and #(ii)# in such a way as to eliminate the electrons, and we get an equation that represents the observed stoichiometric change:

#(i) + 2xx(ii)# gives...............................

#MnO_2(s)+4H^(+)+ 2Cl^(-)rarrMn^(2+)+Cl_2(g) + 2H_2O(l)#

Given the half equation method, for every electron loss, #"oxidation"#, there is thus a FORMAL electron gain, #"reduction".#

The #(s)#, #(g)#, #(l)# represents the physical state of the reactant/product, i.e. #"solid/gas/liquid/"#.........The ions, i.e. #Mn^(2+)# etc. are assumed to be present in solution, and thus do not need a descriptor of state. Happy?