This is an oxidation-reduction (redox) reaction:

6 I-I - 6 e- → 6 I0 (oxidation)

2 CrVI + 6 e- → 2 CrIII (reduction)

NaI is a reducing agent, Na2Cr2O7 is an oxidizing agent.

### Word equation

sodium bichromate + sodium iodide + sulfuric acid → sodium sulfate + iodine +

chromium sulfate + water

### Input interpretation

Na2Cr2O7 + NaI + H2SO4 → Na2SO4 + I2 + Cr2(SO4)3 + H2O

Sodium sodium sulfuric sodium iodine chromium water

bichromate iodide acid sulfate sulfate

### Balanced equation

Balance the chemical equation algebraically:

Na2Cr2O7 + NaI + H2SO4 → Na2SO4 + I2 + Cr2(SO4)3 + H2O

Add stoichiometric coefficients, c, to the reactants and products:

c1Na2Cr2O7 + c2NaI + c3H2SO4 →c4Na2SO4 + c5I2 + c6Cr2(SO4)3 + c6H2O

Set the number of atoms in the reactants equal to the number of atoms in the products for Cr, Na, O, I, H and S:

Cr: 2c1 = 2c6

Na: 2c1 + c2 = 2c4

O: 7c1 + 4 c3 = 4c4 +12c6 +c7

I: c2 = 2c5

H: 2c3 = 2c7

S: c3 = c4 + 3c6

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c1 = 1 and solve the system of equations for the remaining coefficients:

c1 = 1 c2=6 c3 = 7

c4=4 c5 = 3 c6=1

c7 = 7

Substitute the coefficients into the chemical reaction to obtain the balanced equation: