## Redox reaction

*3 Pb0 - 6 e- → 3 PbII (oxidation)*

*6 HI + 6 e- → 6 H0 (reduction)*

In this reaction, Pb acts as the reducing agent, while H3PO4 acts as the oxidizing agent.

## word equation

* phosphoric acid + lead → hydrogen + lead(II) phosphate*

## Input interpretation

* 2 H3PO4 + 3 Pb → 3 H2 + Pb3(PO4)2*

The balanced equation can be obtained by adding stoichiometric coefficients, c, to the reactants and products:

* c1 2H3PO4 + c2 3Pb → c3 3H2 + c4 Pb3(PO4)2*

To balance the equation, we must ensure that the number of atoms in the reactants is equal to the number of atoms in the products for H, O, P, and Pb. We can set an arbitrary value, typically one, for one of the coefficients to keep them small. For instance, we can set c4 = 1 and solve the system of equations for the remaining coefficients:

*c1 = 2*

*c2 = 3*

*c3 = 3*

*c4 = 1*

Substituting the coefficients into the chemical equation gives us the balanced equation:

* 2 H3PO4 + 3 Pb → 3 H2 + Pb3(PO4)2*

In summary, the balanced equation for the redox reaction between lead and phosphoric acid is 2 H3PO4 + 3 Pb → 3 H2 + Pb3(PO4)2, which can be represented by the word equation: phosphoric acid + lead → hydrogen + lead(II) phosphate.