Balancing redox reactions requires a new approach that accounts for the transferring of electrons between reactants. Now we're going to say here that redox reactions not only balance the atoms of elements, but also charge and the number of electrons. To talk about balancing half of redox reactions, we take a look at half reactions. We're going to say here balancing a redox reaction begins with identifying its half reactions, and a half reaction is either the oxidation or reduction reaction portion of a redox reaction. Now we're going to say here usually a half reaction is obtained by identifying the elements that are found as oxygen and hydrogen. So we're going to go step by step in first identifying what half reactions are, and then later on the best way to balance an overall redox reaction.
Balancing Redox Reactions (Simplified) - Online Tutor, Practice Problems & Exam Prep
Balancing Redox Reactions requires a new approach that accounts for the transfer of electrons between reactants.
Balancing Redox Reactions
Balancing Redox Reactions (Simplified) Concept 1
Video transcript
Balancing a redox reaction begins with identifying its half reactions.
Balancing Redox Reactions (Simplified) Example 1
Video transcript
Identify the half reactions from the following redox reaction. We have manganese 5 ion reacting with chloride ion to produce manganese 2 ion and chlorine gas. To give our 2 half reactions, we basically match up elements that are not oxygen and hydrogen. So manganese will be matched up with manganese, chlorine will be matched up with chlorine. So my first half reaction would just be the manganese 5 ion giving directly the manganese 2 ion. And my other half reaction would be the chloride ion, giving me directly the chlorine gas. Here, don't forget the phases that are included, so these would be aqueous, and then here this would be aqueous, and this would be a gas. So we'd have 2 half reactions, one between the manganese ions and one with the chlorine elements.
Balancing Redox Reactions (Simplified) Example 2
Video transcript
So here we're going to balance this redox reaction. Here we have solid nickel and cobalt(III) ions reacting to produce nickel(II) ions and cobalt(II) ions. So, step 1 is we have to break the full redox reaction into two half-reactions. Remember, to do that, we match up elements with elements. So nickel matched with nickel ion. That'd be solid nickel gives me nickel(II) aqueous as one half-reaction, and then cobalt with cobalt. So cobalt(III) ion giving me cobalt(II) ion. Step 2, we balance the overall charge by adding electrons to the more positively charged side of each half-reaction. So what we're going to say here is, if we look, solid nickel here has no charge that we can see. If it did have a charge, it would be present, so its overall charge is 0. On the other side, we have nickel(II) ions, so the overall charge on this side is +2. We add electrons to the more positive side, so the right side, and we add enough electrons so that its overall charge here will match up with the overall charge over on the other side. So I need to go from +2 to 0 in terms of overall charge, that means I need to add 2 electrons. Let's go to the other half-reaction. Here this cobalt ion is +3, so the overall charge on this side is +3. And then here the overall charge is +2. I have to add electrons to the more positive side, so the cobalt(III) ion side, and I add enough electrons so that its charge overall matches the charge overall here on the lesser side. To do that, I'd have to add 1 electron. Alright. So one half-reaction has 2 electrons, the other one has 1. Now here underneath step 2, we say if the number of electrons of both reactions differ, then multiply to get the lowest common multiple. So here this is 2 electrons in this half-reaction, and this one here is 1 electron. So I'd have to multiply this half-reaction by 2 in order for this to become 2 electrons. Lastly, we combine the half-reactions and we cross out the electrons on both sides. Alright. So everything's going to come down. So we're going to have here solid nickel gives me nickel(II) aqueous, plus the 2 electrons. The other half-reaction I multiplied everything by 2, so that's 2 Co(III) aqueous plus 2 electrons gives me 2 Co(II) aqueous. So here the electrons cancel out, everything else comes down, so my balanced equation will be nickel solid + 2 cobalt(III) ions aqueous gives me 1 nickel(II) ion + 2 cobalt(II) ions. So this here would represent my balanced redox reaction.
Balance the following redox reaction.
Problem Transcript
Balance the following redox reaction.
Problem Transcript
Balance the following redox reaction.