This is an **oxidation-reduction** (redox) reaction:

*2 Br-I - 2 e- → 2 Br0 (oxidation)*

*2 HI + 2 e- → 2 H0 (reduction)*

** CaBr2** is a

**reducing**agent,

**is an**

*H2O***oxidizing**agent.

### Word equation

* calcium bromide + water → hydrogen + calcium hydroxide + bromine*

### Input interpretation

*CaBr2 + H2O → H2 + Ca(OH)2 + Br2*

*Calcium water hydrogen calcium bromine*

*Bromide hydroxide*

### Balanced equation

Balance the chemical equation algebraically:

*CaBr2 + H2O → H2 + Ca(OH)2 + Br2*

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

*CaBr2 + H2O → H2 + Ca(OH)2 + Br2*

Set the number of atoms in the reactants equal to the number of atoms in the products for ** Br, Ca, H **and

*O:**Br: 2c1 = 2cs*

*Ca: c1 = c4*

*H: 2c2 = 2c3 + 2c4*

*O: c2 = 2c4*

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=2*

*c3=1 c4=1*

*c5=1*

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

**Answer**:

*CaBr2 + H2O → H2 + Ca(OH)2 + Br2*