How Do You Balance Complex Redox Equations with Multiple Oxidation Changes?

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Balancing complex redox equations involves identifying oxidation states and separating the reaction into half-cells. In the given reaction, chromium (Cr) transitions from +3 to +6, iodine (I) from -1 to +7, and chlorine (Cl) from 0 to +1, indicating that Cr and I lose electrons while Cl gains them. To solve the equation, it is recommended to break it into half-cells, balancing exotic elements like potassium (K), chromium, iodine, and chlorine first. Following standard half-cell balancing procedures will help ensure that both charge and atom counts are correct. Properly balancing each half-cell will lead to a successful overall equation.
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hi all, my chemistry assignment had a bunch of questions on redox equations balancing. i got stuck on this one:

\CrI_{3} + KOH + \Cl_{2} \rightarrow \K_{2}CrO_{4} + \KIO_{4} + KCl + \H_{2}O [\tex]<br /> <br /> so far, i tried to find ON numbers of the various elements.<br /> Cr changes from +3 to +6<br /> I changes from -1 to +7<br /> Cl changes from 0 to +1<br /> <br /> this means that Cr and I undergo a loss of electrons, while Cl undergoes a gain. i don&#039;t know how to solve this question when there are multiple LEO.<br /> <br /> can you help? =)<br /> <br /> [ EDIT : my attempt at LATEX has proved to be yet another disaster ]
 
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For your latex in the final tex it should be a FORWARD SLASH, not a backslash... thus the mess. Also you don't need a backslash to start off a new word. ANyway it look like this:

CrI_{3} + KOH + Cl_{2} \rightarrow K_{2}CrO_{4} + KIO_{4} + KCl + H_{2}O

i suggest you break into half cells as follows
CrI_{3} \rightarrow K_{2}CrO_{4} + KIO_{4}
KOH \rightarrow H_{2}O
Cl_{2} \rightarrow KCL

Balance everything exotic (like K, Cr, I, Cl) first. Then follow the usual steps for half cells. I know it doesn't look like it will work.. but if you balance each half cell properly in temrs of charge and atoms yu should be fine
 
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