In many residential water systems, the aqueous Fe3+ concentration is high enough to stain sinks and turn drinking water light brown. The iron content is analyzed by first reducing the Fe3+ to Fe2+ and then titrating with MnO4- in acidic solution.

Question

In many residential water systems, the aqueous Fe3+ concentration is high enough to stain sinks and turn drinking water light brown. The iron content is analyzed by first reducing the Fe3+ to Fe2+ and then titrating with MnO4- in acidic solution.

Fe2+(aq) + MnO4-(aq) ? Mn2+(aq) + Fe3+(aq)

Balance the skeleton reaction of the titration step. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)

answer: ?????????

In the process of titration iron gets reduced and Mn gets oxidized in presence of acidic condition.

Now if we write redox half cell equation for both the process we got,

Fe3= ---- Fe2+ + e ............. 1st redox equation where it is reduced

MnO-4 + H+ ---- Mn2+ + H2O ...............2nd redox equation where it is oxidized

now from 2nd redox equation we can balance both side to get balanced no of oxygen and hydrogen as per rule of balancing

MnO-4 + H+ -------- Mn+2 + H2O

MnO-4 + 8H+ -------- Mn+2 + 4H2O

Now to balance electrons in both the side

MnO-4 + 8H+ + 5e -------- Mn+2 + 4H2O................3rd redox equation

now by adding both 3rd and 1st redox equation we got

MnO-4 + 8H+ + 5e + Fe+3 -------- Mn+2 + 4H2O..+ Fe

hence to balance both side electrons finally we got the equation as

MnO4- + 8H+ + 5Fe2+ + 5e- -> Mn2+ + 4H2O + 5Fe3+ + 5e-

Canceling out the electrons:

MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+
and this is the balanced redox equation for the above reaction . Right? thank you.•chemistry - DrBob222, Saturday, November 14, 2015 at 11:31am
I can't tell the difference between the question and what I assume is your answer.The skeleton equation is simply
5Fe^2+ + MnO4^- ==> 5Fe^3+ + Mn^2+ and you can add the states.

Most of the rest of what you wrote is not right but the final equation you came up with is correct.
All of that explanation about Fe being reduced and Mn being oxidized is rubbish which I've copied here as
"In the process of titration iron gets reduced and Mn gets oxidized in presence of acidic condition."

I don't know what I did wrong..it said hint: write the oxidation and reduction half reactions. thank you, Dr. BOB.

Answer 1

The oxidation half equation is

Fe^2+ ==> Fe^3+ + e

The reduction half equation is
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O

Answer 2

I apologize for any confusion in my previous response. Based on the information you provided, the skeleton equation for the titration step is:

5Fe^2+(aq) + MnO4^-(aq) -> 5Fe^3+(aq) + Mn^2+(aq)

This equation represents the oxidation of Fe^2+ to Fe^3+ and the reduction of MnO4^- to Mn^2+. The equation is balanced with the lowest possible coefficients and includes the states of matter under SATP conditions.

Answer 3

I apologize for any confusion. To balance the skeleton reaction for the titration step, you need to write the oxidation and reduction half-reactions separately and then combine them to form the balanced overall equation.

The oxidation half-reaction involves the loss of electrons (oxidation) and can be written as:

Fe2+ -> Fe3+ + e-

The reduction half-reaction involves the gain of electrons (reduction) and can be written as:

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Now, you need to balance the number of electrons transferred by multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 1. After balancing the electrons, the two half-reactions can be combined to form the balanced overall equation:

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O

It's important to note that the hint given to you was to write the oxidation and reduction half-reactions, not to provide an explanation of how the oxidation and reduction occur. The explanation you provided in your previous response was not necessary for balancing the equation.