In this reaction, hydrogen (H2) is oxidized to form hydrogen iodide (2 HI) while potassium iodide (2 KI) is reduced to form potassium (2 K0). Hydrogen (H2) acts as the reducing agent while potassium oxide (K2O) acts as the oxidizing agent.

### Word Equation

*potassium oxide + hydrogen → potassium + water*

which can be translated into the chemical equation:

*K2O + H2 → K + H2O*

To balance this equation, we need to add stoichiometric coefficients to each reactant and product. Let's call them c1, c2, c3, and c4 for K2O, H2, K, and H2O, respectively. We can then use the conservation of atoms to set up a system of equations:

*K: 2c1 = c3*

*O: c1 = c4*

*H: 2c2 = 2c4*

Since we have three equations and four variables, we need to set one coefficient arbitrarily. Let's set c1 to 1 and solve for the remaining coefficients:

*c1 = 1, c2 = 1, c3 = 2, c4 = 1*

Substituting these coefficients back into the chemical equation, we get:

*K2O + H2 → 2 K + H2O*

Therefore, the balanced equation is:

*K2O + H2 → 2 K + H2O*

In summary, this redox reaction involves the oxidation of hydrogen and reduction of potassium iodide. By balancing the equation, we can see that potassium oxide reacts with hydrogen to form potassium and water.