Fe2+ + Cr2o72- = Fe3+ + Cr3+ how do I balance this reaction in a acidic solution?

Answer 1

There are two methods for balancing the redox reactions.
(1) Oxidation number method
(2) Half-reaction method

Half-reaction method, although a little bit time consuming, is easier for students who get difficulty in determining the oxidation number of the species involved in the equation. In your question there is only one substance, dichromate ion Cr2O7^2-, to cause a difficulty, others are in the form of monoatomic ions. Therefore, it will be more convenient to use oxidation number method. ( May be I can explain you the half-reaction method later in your another question).

(1) Determine the oxidation numbers of all species.
Cr2O7^2- ion. Since the oxidation number of O is -2 in most cases (except peroxides, superoxides, and OF2), 7 O atoms have a charge of 7(-2) = -14. -12 of this charge must be neutralized by 2 Cr atoms to leave -2 charge for the dichromate ion. Therefore, 2 Cr atom must possess +12 charge. Then, the oxidation number of single Cr atom should be +6.

(2) Determine the substance oxidized and the substance reduced.
Fe is the sustance oxidized (from +2 to +3)
Cr is the sustance reduced (from +6 to +3)

(3) Determine the electrons to be lost and gained. Since Fe is oxidized, it loses electron, since Cr is reduced it gains electron.

Fe^2+ ------------> Fe^3+ + 1e-
2Cr^6+ + 6e- ----------> 2Cr^3+

Note that Cr in the reactant side is 2 moles and in the product side is 1 mole. It is better to balance them first before calculating the number of e-. Because in some equations this creates a problem in balancing.

(4) Equate the number of electrons by multiplying the half equations by proper coefficents. Since in your question, substance oxidized loses 1 e- and substance reduced gains 6e-, it is sufficent to multiply the first equation by 6.

6Fe^2+ ------------> 6Fe^3+ + 6e-
2Cr^6+ + 6e- ----------> 2Cr^3+

(5) Add two half reactions:
Electrons lost will cancel electrons gained.

6Fe^2+ + 2Cr^6+ --------> 6Fe^3+ + 2Cr^3+

The coefficents of the substances show that how many moles of the related substances will be in the main equation.

(6) Carry the coefficents to the main equation.
There must be 6Fe^2+ and 2Cr^6+ in the left side of the equation. Did you notice we have already 2 Cr in the left side. If we considered the reduction of 1 mole Cr atom, this would cause a problem.
There must be 6Fe^3+ and 2Cr^3+ in the right side of the equation.
After carrying the coefficents we obtain;

6Fe^2+ + Cr2O7^2- --------> 6Fe^3+ + 2Cr^3+

(7) The next step is the charge balance. This is done by H^+ ion in acidic solutions and by OH^- in basic solutions.

The charge on the left: +12 - 2 = +10
The charge on the right: +18 +6 = +24

To maintain the charge balance, +14 charge is necessary to the left side. This is done by adding 14H^+ ion.

6Fe^2+ + Cr2O7^2- + 14H^+ --------> 6Fe^3+ + 2Cr^3+

(8) The last step is to balance the number of O atoms by adding H2O.
There are 7 O atom on the left, therefore we have to add 7 H2O to the right. This also balance 14 H atom.

The balanced equation:
6Fe^2+ + Cr2O7^2- + 14H^+ --------> 6Fe^3+ + 2Cr^3+ + 7H2O

Answer 2

Fe2+ goes from +2 to +3
Cr goes from +6 to +3
The balanced equation is
6 Fe2+ + Cr2O72- + 14 H+ --> 6 Fe3+ + 2Cr3+ + 7H2O

Answer 3

Yes; but HOW?

Balance main element
Balance oxygens using water
Balance hydrogens using hydrogen ions
Balance charge using electrons
Recheck that everything is balanced, including charge
Do this for both the oxidation and reduction, cross multiply by a number of electrons so they will cancel out, combine and if possible simplify.

Memorise this drill, this is one of the things you are expected to be able to do It always works.

In this instance,

Cr2O7 2- = 2 Cr3+
Cr2O7 2- = 2 Cr3+ + 7H2O
Cr2O7 2- + 14 H+ = 2 Cr3+ + 7 H2O
Cr2O7 2- + 14H+ + 6 e- = 2 Cr3+ + 7H2O

Answer 4

B/ Cr2O72- + Fe2+ → Cr3+ + Fe3+ (Acidic soln.)