In: Chemistry

Balance the **redox** reaction

FeSO_{4} + H_{2}SO_{4} +
KMnO_{4} ---> Fe_{2}(SO_{4})_{3}
+ K_{2}SO + MnSO_{4} + H_{2}O

What is the coefficient of FeSO_{4} in the balanced
equation? **Please show work.**

Above reaction is an example for a "redox" reaction.

Here, Mn is an oxidizing agent and it undergo reduction (i.e. lowering of charge from 7+ to 2+ state) and Fe is a reducing agent (i.e. increasing of charge from 2+ to 3+ state).

In order to understand the reaction more appropriately, look at
its ionic equation.

K^{+} + MnO_{4}^{-} + Fe^{2+} +
SO_{4}^{2-} + 2H^{+} +
SO_{4}^{2-}
2Fe^{3+} + 3SO_{4}^{2-} + Mn^{2+} +
SO_{4}^{2-} + 2K^{+} +
2SO_{4}^{2-} + H_{2}O

Next, rewrite the reaction without spectator ions

MnO_{4}^{-} + Fe^{2+}
Mn^{2+} + Fe^{3+}

In order to balance a redox problem, split the reaction equation in
to an oxidation and reduction half-equations.

Further, balance everything in the equations except of H and O,
balance O by adding H_{2}O to the opposite side, balance H
by adding H^{+} to the opposite side. Also, add
e^{-} to balance charge.

Finally, multiply the half-equations by a common term to get
e^{-} lost equal to e^{-} gained.

Oxidation:

Fe^{2+}
Fe^{3+} + e^{-}

Reduction:

MnO_{4}^{-}
Mn^{2+} (note: Mn on left side is 7+)

MnO_{4}^{-}
Mn_{2}^{+} + 4H_{2}O

8H^{+} + MnO_{4}^{-}
Mn^{2+} + 4H_{2}O

5e^{-} + 8H^{+} +
MnO_{4}^{-}
Mn^{2+} + 4H_{2}O

The two half-reactions:

Fe^{2+}
Fe^{3+} + e^{-}

5e^{-} + 8H^{+} +
MnO_{4}^{-}
Mn^{2+} + 4H_{2}O

To balancing e^{-} multiply by factor 5

(Fe^{2+}
Fe^{3+} + e^{-}) x 5

5Fe^{2+}
5Fe^{3+} + 5e^{-}

5e^{-} + 8H^{+} +
MnO_{4}^{-}
Mn^{2+} + 4H_{2}O

Adding back together:

5Fe^{2+} + 5e^{-} + 8H^{+} +
MnO_{4}^{-}
5Fe^{3+} + 5e^{-} + Mn^{2+} +
4H_{2}O

Upon Simplification we will get balanced net ionic equation

5Fe^{2+} + 8H^{+} +
MnO_{4}^{-}
5Fe^{3+} + Mn^{2+} + 4H_{2}O

Then, add back K^{+} and SO_{4}^{2-}

4H_{2}SO_{4} + KMnO_{4} + 5
FeSO_{4}MnSO_{4}
+ 5/2 Fe_{2}(SO_{4})_{3} + K^{+} +
1/2 SO_{4}^{2-} + 4H_{2}O

Multiply the above equation by factor 2 in order to eliminate the fractional coefficients.

8H_{2}SO_{4} + 2KMnO_{4} + 10
FeSO_{4}
2MnSO_{4} + 5 Fe_{2}(SO_{4})_{3} +
K_{2}SO_{4} + 8H_{2}O (Balanced)

Therefore, **10** is the coefficient of
FeSO_{4} in the balanced equation.

9) for the reaction
[ 2Al + 3H2SO4 → Al2(SO4)3 + 3 H2SO4] 6.0 g of Al (MW=26.98 g/mol)
are added to 12.0 mL of 0.20 M of H2SO4
A) what is the limiting reagent in the above
reaction?
b) how many miles of H2SO4 will be formed?
c) how many grams of Al2(SO4)3 (MW=342.15 g/mol) are
produced in the reaction?

Balance the following redox reaction if it occurs in
H2SO4. What are the coefficients in front of
H2O and Cr2(SO4)3 in
the balanced reaction?
C3H8O2(aq) +
K2Cr2O7(aq) →
C3H4O4(aq) +
Cr2(SO4)3(aq)
Can someone please explain everything step by step since how to
determine oxidation numbers and half reactions?

Balance the following redox reaction using the half reaction
method:
Fe2+ + MnO4 1- -> Fe3+ + Mn2- (acidic solution)

show the steps to balance the following equations
Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3
and
C6H10 + O2 = CO2 + H2O

Balance the following equation, remove all mole fractions
C5H12O +
K2Cr2O7 +
H2SO4 =
Cr2(SO4)3 +
C2H4O2 +
K2SO4 + H2O
I need the half reaction 1, half reaction 2, and the complete
balanced equation (with K+ and
SO42-)
Please explain your steps, I'm trying to study for a test and
I'm not sure how to go about this. Thank you!

In a particular redox reaction, MnO2 is oxidized to MnO4– and
Fe3 is reduced to Fe2 . Complete and balance the equation for this
reaction in acidic solution. Phases are optional.

Potassium permanganate, KMnO4, is a powerful oxidizing agent.
The products of a given redox reaction with the permanganate ion
depend on the reaction conditions used. In basic solution, the
following equation represents the reaction of this ion with a
solution containing sodium cyanide:
MnO4−(aq)+CN−(aq)→MnO2(s)+CNO−(aq) Since this reaction takes place
in basic solution, H2O(l) and OH−(aq) will be shown in the
reaction. Places for these species are indicated by the blanks in
the following restatement of the equation: MnO4−(aq)+CN−(aq)+
−−−→MnO2(s)+CNO−(aq)+ −−−...

Standardize the KMnO4 using 0.1M (NH4)2(FE(SO4)26H2O (remember
dissolve in H2SO4)
please display your full workings

2. Balance the first redox reaction in acidic solution and the
second redox reaction in basic solution.
S + NO3- → SO2 + NO
Cr2O72- + Fe2+ -----> Cr3+ + Fe3+

For the following reaction:
3H2SO4(g) + 2Fe(OH)3(s) →
Fe2(SO4)3(s) +
6H2O(l)
1. Calculate ΔH°rx (in kJ) for this reaction. Report
your answer to two decimal places in standard notation
(i.e. 123.45 kJ).
2. Calculate ΔS°rx (in J/K) for this reaction. Report
your answer to two decimal places in standard notation
(i.e. 123.45 J/K).
3. Calculate ΔG°rx (in kJ) at 511.71 K for this
reaction. Report your answer to two decimal places in standard
notation (i.e. 123.45 kJ).
Assume ΔH°f and...

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