Question #17435

Jan 13, 2016

$3 U {O}^{2 +} + C {r}_{2} {O}_{7}^{2 -} + 8 {H}^{+} \to 3 U {O}_{2}^{2 +} + 2 C {r}^{3 +} + 4 {H}_{2} O$

Explanation:

This is a redox reaction:

$U {O}^{2 +} \left(a q\right) + C {r}_{2} {O}_{7}^{2 -} \left(a q\right) \to U {O}_{2}^{2 +} \left(a q\right) + C {r}^{3 +} \left(a q\right)$

The oxidizing agent is the $C {r}_{2} {O}_{7}^{2 -}$ and the reducing agent is the $U {O}^{2 +}$.

$U$ is getting oxidized since its oxidation number is increasing from $+ 4$ to $+ 6$.

$C r$ is getting reduced since its oxidation number is decreasing from $+ 6$ to $+ 3$.

To balance this reaction we will split it into two half equations and we will assume that is happening in acidic medium:

Oxidation: $\left(U {O}^{2 +} + {H}_{2} O \to U {O}_{2}^{2 +} + 2 {H}^{+} + 2 {e}^{-}\right) \textcolor{red}{\times 3}$

Reduction: $C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \to 2 C {r}^{3 +} + 7 {H}_{2} O$

RedOx: $\textcolor{red}{3} U {O}^{2 +} + C {r}_{2} {O}_{7}^{2 -} + \textcolor{b l u e}{8} {H}^{+} \to \textcolor{red}{3} U {O}_{2}^{2 +} + \textcolor{g r e e n}{2} C {r}^{3 +} + \textcolor{b l u e}{4} {H}_{2} O$

Here is a video that explains how to balance RedOx reactions in Acidic Medium:
Balancing Redox Reactions | Acidic Medium.

Basic Medium:
Balancing Redox Reactions | Basic Medium.