Question #17435

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Question #17435

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Answer 1 · 2016-01-13T20:49:14

$3 U O^{2 +} + C r_{2} O_{7}^{2 -} + 8 H^{+} \to 3 U O_{2}^{2 +} + 2 C r^{3 +} + 4 H_{2} O$ $3UO^(2+)+Cr_2O_7^(2-)+8H^(+)->3UO_2^(2+)+2Cr^(3+)+4H_2O$

Explanation:

This is a redox reaction:

$U O^{2 +} (a q) + C r_{2} O_{7}^{2 -} (a q) \to U O_{2}^{2 +} (a q) + C r^{3 +} (a q)$ $UO^(2+)(aq)+Cr_2O_7^(2-)(aq)->UO_2^(2+)(aq)+Cr^(3+)(aq)$

The oxidizing agent is the $C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ and the reducing agent is the $U O^{2 +}$ $UO^(2+)$ .

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

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

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

Oxidation: $(U O^{2 +} + H_{2} O \to U O_{2}^{2 +} + 2 H^{+} + 2 e^{-}) \times 3$ $(UO^(2+)+H_2O->UO_2^(2+)+2H^++2e^-)color(red)(xx3)$

Reduction: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ $Cr_2O_7^(2-)+14H^(+)+6e^(-)->2Cr^(3+)+7H_2O$

RedOx: $3 U O^{2 +} + C r_{2} O_{7}^{2 -} + 8 H^{+} \to 3 U O_{2}^{2 +} + 2 C r^{3 +} + 4 H_{2} O$ $color(red)(3)UO^(2+)+Cr_2O_7^(2-)+color(blue)8H^(+)->color(red)(3)UO_2^(2+)+color(green)(2)Cr^(3+)+color(blue)(4)H_2O$

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.

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Question #17435

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