How do you represent the oxidation of #Cu^+# by #MnO_4^(-)#?

1 Answer
Nov 30, 2015

Answer:

#Cu^(+) rarr Cu^(2+) + e^-# OXIDATION
#MnO_4^(-) + 8H^(+) + 5e^(-)rarr Mn^(2+) + 4H_2O# REDUCTION

Explanation:

We can write the overall redox equation by adding #5# #xx# the oxidation equation to #1# of the reduction reaction to give:

#5Cu^+ + MnO_4^(-) + 8H^(+) rarr Mn^(2+) + 5Cu^(2+) + 4H_2O(l)#

This reaction is stoichiometrically balanced with respect to mass and charge, and it is self indicating. Why? Because permanganate ion, #MnO_4^-#, has a strong purple colour, whereas #Mn^(2+)# is almost colourless. The colour of the permanganate ion will persist at the endpoint.

Concentration of cuprous ion: #(n_(Cu^+))/(2.49xx10^-2*L)#

Moles of permanganate ion #=# #0.128*mol*cancel(L)^(-1)xx3.19xx10^(-2)cancel(L)# #=# #4.08xx10^(-3) mol#. Given the stoichiometry there were #5xx4.08xx10^(-3)mol# #=# #2.042xx10^(-2)# #mol# of cuprous ion.

SO the concentration of the original solution was #(2.042xx10^(-2)*mol)/(2.49xx10^-2*L)# #=# #0.820# #mol*L^(-1)# in #Cu^+# ion.