# Which element is oxidized in this reaction? 2CuO+C→2Cu+CO2 Which element is reduced in this reaction? 16H++2Cr2O72−+C2H5OH→4Cr3++11H2O+2CO2

Jun 11, 2017

The element oxidized is the one whose oxidation number INCREASES......

#### Explanation:

And thus......

$\stackrel{+ I I}{\text{ 2CuO}} + C \rightarrow 2 \stackrel{0}{C} u + C {O}_{2} \uparrow$

And here while copper (as $C {u}^{2 +}$) is reduced, carbon is oxidized, $\stackrel{0}{C} \rightarrow \stackrel{+ I V}{C}$.

The second reaction represents the reduction $C r \left(V I +\right)$ to $C r \left(I I I +\right)$

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$ $\left(i i\right)$

And the oxidation of ethanol to carbon dioxide:

${H}_{3} C - C {H}_{2} O H + {H}_{2} O \rightarrow 2 C {O}_{2} + 8 {H}^{+} + 8 {e}^{-}$ $\left(i i i\right)$

i.e. ${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H \rightarrow 2 \stackrel{+ I V}{C} {O}_{2}$

We takes $4 \times \left(i i\right) + 3 \times \left(i i i\right)$ to eliminate the electrons.....

$3 {H}_{3} \text{CC} {H}_{2} O H + 3 {H}_{2} O + 4 C {r}_{2} {O}_{7}^{2 -} + 56 {H}^{+} + 24 {e}^{-} \rightarrow 8 C {r}^{3 +} + 28 {H}_{2} O + 6 C {O}_{2} + 24 {H}^{+} + 24 {e}^{-}$

To give finally............

$3 {H}_{3} \text{CC} {H}_{2} O H + 4 C {r}_{2} {O}_{7}^{2 -} + 32 {H}^{+} \rightarrow 8 C {r}^{3 +} + 25 {H}_{2} O + 6 C {O}_{2}$

$\text{Dichromate ion}$ is reduced to $\text{chromic ion}$.........