Question

Match the following given the reaction below:
$CuO+H_2⟶Cu+H_{2}O$

• A. CuO
• B. $H_2$
• C. Cu
• D. $H_{2}O$

Answer 1

$\stackrel{+2}{\mathrm{Cu}}\stackrel{-2}{O}+{\stackrel{0}{H}}_2\to \stackrel{0}{\mathrm{Cu}}+\stackrel{+1}{H_2}\stackrel{-2}{O}$

• Oxidation state of copper (Cu) decreases from +2 to zero so copper (Cu) is reduced species.
• Oxidation state of hydrogen in $H_{2}O$ is +1 which increased from zero in $H_2$. So, $H_{2}O$ is the oxidized species.
• CuO is the oxidizing agent because the oxidation state of copper decreases as the product - Cu is formed.
• $H_2$ is the reducing agent because the oxidation state of hydrogen increases as the product $H_{2}O$ is formed