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."
It daid; 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."
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."
Dr. BOB:
It said Try again: This is an oxidation-reduction reaction in which electrons are transferred from one substance to another substance. Identify the substances involved in the oxidation half-reaction and the reduction half-reaction. Mass balance each half reaction by adding H2O to balance oxygen atoms and H+ to balance hydrogen atoms. Hint: Write the oxidation and reduction half reactions.
Help. thank you.
Sorry....I wrote it so many times on accident.
To balance the skeleton reaction for the titration step, we need to identify the substances involved in the oxidation half-reaction and the reduction half-reaction.
Let's start with the oxidation half-reaction:
Fe2+(aq) is being oxidized to Fe3+(aq). To balance this half-reaction, we need to add H2O to balance the oxygen atoms and H+ to balance the hydrogen atoms.
Fe2+(aq) -> Fe3+(aq) + e-
Next, let's move to the reduction half-reaction:
MnO4-(aq) is being reduced to Mn2+(aq). Again, we need to add H2O and H+ to balance the atoms.
MnO4-(aq) + 8H+(aq) -> Mn2+(aq) + 4H2O(l)
Now, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 1 to equalize the number of electrons transferred.
5Fe2+(aq) -> 5Fe3+(aq) + 5e-
MnO4-(aq) + 8H+(aq) -> Mn2+(aq) + 4H2O(l)
Finally, we can combine both half-reactions to obtain the balanced equation:
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) -> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
And this is the balanced equation for the titration step.