How to balance Cr2O7^-2 + (H+) + I- to give Cr^+3 + H2O +I3^-1
Answers
Answer:
Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.
Cr2O72- + H+ + I- → Cr3+ + I2 + H2O
Step 2. Separate the redox reaction into half-reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.
a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).
Cr+62O-272- + H+1+ + I-1- → Cr+33+ + I02 + H+12O-2
b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down).
O:I-1- → I02(I)
R:Cr+62O-272- → Cr+33+(Cr)
c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).
O:I-1- → I02
R:Cr+62O-272- → Cr+33+
Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas. Never change a formula when balancing an equation. Balance each half reaction separately.
a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.
Answer:
Cr2O7{2-} + 6 I{-} + 14 H{+} = 2 Cr{3+} + 3 I2 + 7 H2O
Explanation: