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:

**Answer:**

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

* *