Question #82f6e

May 17, 2017

$K$ is being oxidized,
$C u$ is being reduced
Since oxidation is the gain in oxidation state (loss of electrons), $K$ is being oxidized because it has on oxidation state of $0$ originally, and as the reaction proceeds, its oxidation state is raised to $+ 1$.
Since reduction is the lowering in oxidation state (gain of electrons), $C u$ is being reduced because it has on oxidation state of $+ 2$ originally, and as the reaction proceeds, its oxidation state is lowered to $0$ from becoming pure $C u$.