Question

How do chromium metal, and `Pb^(2+)` interact electrochemically?

Answer 1

Chromium metal is oxidized to `Cr^(3+)`, and lead is reduced from `Pb^(2+)` to `Pb^(0)`. Half equations follow.

Explanation:

Oxidation half equation:

`Cr rarr Cr^(3+) + 3e^(-) (i)`

Reduction half equation:

`Pb(NO_3)_2 + 2e^(-) rarr Pb^0 + 2NO_3^-` `(ii)`

Both equations are balanced with respect to mass, and to charge. The overall equation is simply the sum of equations (i) and (ii) such that electrons do not appear in the final reaction: i.e. `2 xx (i) + 3 xx (ii):`

`2Cr + 3Pb(NO_3)_2 rarr 2Cr^(3+) + 3Pb^0 + 6NO_3^-`.

Note that the overall redox equation is balanced with respect to mass and charge, as indeed it must be.