**Concept:** Balancing Redox Reactions

Hey guys. We're going to continue with our discussion of redox reactions by now learning how to balance them whether they're in acidic solutions or basic solutions. So, we know the ins and outs of redox reactions, we know oxidation, we know the reduction from the past videos, but now if our goal is to learn how to balance then when they're in acidic solutions or basic solutions and to do that we're going to follow a different set of rules, they almost go hand-in-hand together except when we get to a basic solution we have to add an additional step. So, first let's to balance it out in an acidic solution. So, here going to say step one, we have to write the equation, the full redox equation first into two half reactions and we'll see exactly how do we do this but basically we look for elements different from oxygen or hydrogen first on either sides of the arrows and bring those down, separating them into two separate equations, once we do that we balance out these elements that are not oxygen or hydrogen's first, once that's done, we can move on to step three, we have to balance our oxygens by adding molecules of water to the opposite side so that both sides have the same number of oxygen, what we do next after that is we have to balance out hydrogen by adding H plus. Now, here comes the tricky part after, we do this, we have to balance out the overall charge of that half reaction by adding electrons to the more positive side and we have to make sure that when we do this, the electrons in both half reactions must match and if they don't, that means we're going to have to multiply either one of our half reactions or maybe even both. So, we're going to say electrons from both half-reactions must match, if not that means we have to balance either one of half reactions or both of them again. Alright, after we do that, once our electrons match up in both half reactions, we can then combine those two half reactions back into our full redox reaction and this time we're going to cross out intermediates and remember intermediates are species that look alike except one will be a reactant and one will be a product, those will cancel out member, if they're on the same side, let's say they're both reactants of products, they don't cancel out at all they actually add up together. So, remember intermediates ones are reactant ones are product, they look alike they'll cancel out, once we've done this an acidic solution, basic solution is basically almost all the same rules except for one additional rule at the end. So, for imbalancing in basic solutions, we follow rules, and once we've accomplished this, we're going to, we're going to balance H plus by adding OH minus ions to both sides of the chemical reaction, when we get to basic solutions, we'll see quickly, what this really means. So, remember it's imperative that you guys first remember the basic rules that we went over, because this tells us what's being reduced and what's being oxidized, remember if you've been reduced you're the oxidizing agent but if you've been oxidized you're the reducing agent, mastering that helps you to move on to this aspect of redox reactions whereas to balance them out and remember you have to remember all the guidelines all the steps for either acidic or basic in order to get the full credit for that question. So, hopefully you guys can remember these rules and will apply them on the next series of videos on how to balance redox reactions.

**Concept:** Balancing Acidic Redox Reactions

Hey guys. In this new video we're going to put to practice some of the rules that we learn to balance redox reactions in acidic solution. So, let's take a look at this first one, here we need to balance out this redox reaction in an acidic solution. So, let's go over the rules that we know, first we're going to break this up into two half reactions. So, we look for the elements different from oxygen or hydrogen first, here we have an N here, here we have another N right here. So, we're going to do is we're going to bring them both down and say that represents our first half reaction. So, we're going to say NO2- gives us NO3-, here we have Mn and, here we have Mn. So, those will also give us another half reaction MnO4- gives us Mn2+. Alright, next we have to balance out elements different from oxygen or hydrogen. So, what we need to see is on the first half reaction, we only have one nitrogen. So, we don't have to worry about balancing they're already balanced, on the other side we have one manganese on each side. So, we don't need to balance that either. Now, we move on to balancing out oxygens by adding water, on this side we have two oxygens, on this side we have three. So, we need to put one mole of water, or one, yeah one mole of water on the left side. So, now that both sides have three, let's look at the other half reaction, here this has four but the other side has none. So, we have to add four moles of water. Now, we balanced out oxygen. Now, it's time to balance out H+. So, looking back on the left half reaction, we're going to say we have two H's right here, but we have none on the product side. So, we have to add two H+ to this side, then looking at the other half reaction, we have 4 times 2, we have 8 H's there. So, we have to add eight H+ here. Now we, this is the tricky part, we have to balance that overall charge. So, we're going to say here is we're going to say water is neutral, there's no charge on this. So, we're going to say it's charge is 0, but nitrite ion NO2- has a negative one charge. So, we're going to say overall this side is negative one, let's look at the other side, here we have H plus for H, but there's two of them so that really counts as plus 2 and then we're going to say our nitrate ion NO3- has a minus one charge. So, that's minus 1. So, overall the charge on this side is plus 1, remember, we have to add electrons to the more positive side so that both sides have the same exact charge. So, we're going to add electrons to the plus 1 side.

Now, we need that plus 1 to become a negative 1 just like on the left side. So, how many electrons do we need to add? we need to add two electrons, because adding two electrons is equivalent to adding a negative 2. So, here we're going to say plus 1 minus 2 equals minus 1. So, both sides are now negative 1, let's go to the other side, here we have eight plusses. So, that's plus 8, we have a negative one here. So, that's minus 1. So, this side is plus 7, here on the other side water again is neutral has no charge. So, it's 0, here we have plus two. So, this side is plus 2 overall. Now, we need to add electrons to the more positive side. So, we're going to add them to the left side because it's plus 7, we need both sides to be the same exact charge. So, how many electrons do I need to add two plus seven so that drops down to plus two, we're going to have to add five electrons, because remember, each electron is negative. So, adding five electrons is equivalent to doing negative 5 plus 7 minus 5 equals plus 2, both sides are now plus 2. Now, finally, we have to check to make sure our electrons are equal, here we have two electrons, but here we have five, they're not the same number. So, we're going to do here is we have to multiply both half reactions by a number so that they both have equal number of electrons so the common multiple between two and five is 10. So, we multiply this whole thing here, times 5 and we multiply this whole thing times 1, and remember, we're doing this because we need to have the same number of electrons on in both half reactions, if we don't then it can't be a valid way of balancing it in acidic solution. Alright, so everything gets multiplied by five. So, we're going to have five NO2- plus 5 waters, gives me 5 and a 3- plus 10 H+ plus 10 electrons and what you should remember here, just remember this for later on, we have electrons as products, if you have electrons as products then this is an oxidation. So, remember, if you have electron this product it's an oxidation, let's look at the other one, everyone is getting multiplied by two. So, this would be 2 MnO4- plus 8 times 2, 16 H+, plus 10 electrons gives me 2 Mn2+, plus 8 H2O, and look we have electrons in the second one as reactants. So, electrons are reactance, reactance means it's reduction, okay? So, remember, if your electrons are products it's oxidation, but if your electrons are reactants its reduction. Alright, so we have to make sure that the electrons cancel out and they do, one's a product ones are reacting, they're intermediates, let's see who else can be intermediates, we can say that here, these 10 H+ that are products, all of them get canceled out by 10 from here, leaving us with 6 left, we're going to say all 5 of these waters that are reactants get cancelled out by 5 from here, leaving us with 3 left, and it looks like those are our only intermediates that we have, everything left we bring it down. So, what we have left at the end, we're going to have 2 MnO4- plus 5 NO2- plus 6 H+, gives me 5 NO3- plus 2Mn 2+ plus 3H2O, so that would be our final answer when balancing this redox reaction in acidic solution, I know it's a lot of moves, a lot of steps to remember, but remember, go at it slowly if you're getting lost, I think the worst part of this whole thing is really looking at the overall charge, but just remember, if you don't see the compound with any type of charge it's 0, if it has a coefficient in front of it that would multiply the charge, whatever it is, making it bigger, and as long as you can get those basics down you'll know how to add the number of electrons you need. So, both sides have the same overall charge and again remember, your electrons in both half reactions must match up, if they don't then it's done incorrectly. Now, that we've done this one I want you guys to attempt to do the next one on your own. So, it's the same basic concept. So, we're going over the same rules, break it up into half reactions first and from there we take it on to do the rest of the steps to balance it in acidic solution, if you get lost it's, okay? Just click on the explanation button and a video of me will show you how best to approach this next problem, good luck guys.

**Problem:** Balance the following reaction in an **acidic** solution.

Cl_{2}(g) + S_{2}O_{3}^{2-}(aq) ----> Cl^{-}(aq) + SO_{4}^{2-}(aq)

**Concept:** Balancing Basic Redox Reactions

Hey guys in this new video we're finally going to look at how do we balance the redox reaction in a basic solution. So, let's take a look at the first example. So, here we're going to have to balance out this massive reaction in a basic solution. So, let's just follow the rules that we know first, when following an acidic solution because remember, the first six rules are the same for basic solutions. So, here we have Mo and here, we have Mo. So, let's bring that down as its own half reaction and here, we have Br and here, we have Br, let's bring those down as their own half reaction next, we have to make sure we balance out elements that are different from oxygen or hydrogen. So, here we have three Mo's but over here we only have one. So, we're going to throw a three in front of that and then on the other side both bromines are just one. So, we don't have to worry about anything there, next we have to balance out oxygens by adding water, on this side, we have nine oxygens on the other side we have none. So, we have to add nine waters here we have four oxygens but on the other side we have none. So, we're going to add four waters, next we're going to balance out H plus. So, here we have nine times two that gives us 18. So, we're going to put 18 H plus on this side and then here, we have 4 times 2, that's 8. So, we have to add 8 h plus to this side. Now, for the tricky part overall charge. Now, the overall charge here, we have 18 times plus 1. So, that's plus 18, minus 3, it'll be plus 15 overall on this side, on the other side Mo and water are both neutral, we don't show any charges for them. So, overall it's 0 on this side then on the other side we have, water is neutral again, but we have a negative with the Br. So, this side is negative 1 overall, then we're going to say, we have 8 times plus 1. So, that's plus 8 and then here this is a minus 2. So, here this is plus 6 on this side. So, we need to balance out the overall charge by adding electrons to the more positive side. So, on this side, we're going to add 15 electrons so that both sides are 0 and then here, plus 6, we're going to add how many electrons we need to get to negative 1, we need to add 7 electrons. Now, we're going to say the electrons definitely don't match. So, we have to think of a number we can multiply them both by so that they both match. So, we have to do here is we have to multiply this 1 by 7 and we multiply this 1 by 15 because the common number between them is 105. So, now we're going to have really big numbers we have to deal with. So, we're going to have 7 Mo3 O9 3 minus plus, we have 18 times 7. So, it's 126 h plus, plus 105 electrons, gives me 45 Mo plus, we're going to have 15 I'm actually, sorry, we're multiplying by 7, I was looking at the 15. So, it's 7 times 3, which is 21 and then we have 7 times 9, which is 63, okay? So, remember to multiplied by 7, I was looking at the 15 for a moment but it's times 7. Now, we look at the other one. So, here we are going to have 15 multiplying every one. So, we're going to have 15 Br minus plus 60 H2O gives me 15 Br O4 2 minus, plus, we're going to have 15 times 8, which is 120 plus 105 electrons. So, there's a lot of big numbers going on here.

Now, we have to cross on intermediates. So, who looks like but ones are reacting one is a product, all 120 of these h plus cancel out with 120 from here leaving us with 6, all 105 electrons of course have to cancel out, all 60 of these waters cancel out with 60 from here, leaving us with 3, bring down everything. So, at the end we're just going to have a balanced redox reaction in an acidic solution but we need it in a basic solution. So, we have to do one additional step. So, we said we have to balance out H plus by adding OH minus to both sides. So, how many H pluses do we have, we have 6 here. So, that means we have to add 6 OH minus here, plus 6 OH minus here and remember, what happens when you have H plus and OH minus together, they're opposite charges so they're going to combine to give us water. So, 6OH minus 6H plus combined to give us six waters but we can have water on both sides of the equation. So, what's going to happen here they're going to cancel each other out. So, all three of these would cancel out with 3 from here, leaving us with three waters. So, we'd say the final answer, we'll just bring down everyone, okay? So, we get this for our final answer, so that last step is a little bit different from what we're custom to seeing but that's the last step you'd have to use in order to bounce it in a basic solution. So, look at how many H+ you have left add OH- to both sides equal to that number. Remember, H+ OH- together gives us water and then you have to cancel out the waters from both sides of the reactant and products, whoever is left is whoever's left, at the end nothing happens to this OH- on this side. So, we just bring it down, so that would be our final answer. Now, that you guys have seen this I want you guys to attempt the next one and I'll give you guys some help on the last one because it's different than all the others, here the common element found in both is xenon. So, I'll actually help you out here, when it's a xenon. So, XeO2 gives me H2Xe and XeO2 gives me XeO4, since xenon is found in both xenon will be part of both half reactions xenon O2. So, now that I set that up for you just follow the rules that we've learned from all the other examples in order to balance it first in an acidic solution, soak it in a basic solution, good luck guys.

**Problem:** Balance the following reaction in a **basic** solution.

XeO_{2} (aq) ----> H_{2}Xe (aq) + XeO_{4} (aq)

Balance the following redox reaction by inserting the appropriate coefficients.

HNO_{3} + H_{2}S → NO + S + H_{2}O

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Balance each of the following redox reactions occurring in basic solution.

a. H_{2}O_{2}(aq) + ClO_{2}(aq) → ClO_{2}^{-}(aq) +O_{2}(g)

b. Al(s)+MnO_{4}^{-}(aq) → MnO_{2}(s) + Al(OH)_{4}^{-}(aq)

c. Cl_{2}(g) → Cl^{−}(aq)+ ClO^{−}(aq)

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Balance the following redox reactions occurring in basic solution.

NO_{2}^{−}(aq) + Al(s) → NH _{3}(g) + AlO_{2}^{−}(aq)

a. NO_{2}^{−}(aq) + H_{2}O(l) + 2Al(s) + OH^{−}(aq) → NH_{3}(g) + 2AlO_{2}−(aq)

b. NO_{2}^{−}(aq) + H_{2}O(l) + 2Al(s) + OH^{−}(aq) → 2NH_{3}(g) + 2AlO_{2}^{−}(aq)

c. NO_{2}^{−}(aq) + H_{2}O(l) + Al(s) + 2OH^{−}(aq) → 2NH_{3}(g) + AlO_{2}^{−}(aq)

d. NO_{2}^{−}(aq) + 2H_{2}O(l) + Al(s) + 2OH^{−}(aq) → NH_{3}(g) + AlO_{2}^{−}(aq)

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Balance the following redox reactions occurring in basic solution.

MnO_{4}^{−}(aq) + Br ^{-}(aq) → MnO_{2}(s) + BrO_{3}^{−}(aq)

a. MnO_{4}^{−}(aq) + H_{2}O(l) + Br^{−}(aq) → MnO_{2}(s) + 2OH^{−}(aq) + BrO_{3}^{−}(aq)

b. 2MnO_{4}^{−}(aq) + H_{2}O(l) + Br^{−}(aq) → 2MnO_{2}(s) + 2OH^{−}(aq) + BrO_{3}^{−}(aq)

c. 2MnO_{4}^{−}(aq) + H_{2}O(l) + Br^{−}(aq) → 2MnO_{2}(s) + OH^{−}(aq) + BrO_{3}^{−}(aq)

d. MnO_{4}^{−}(aq) + H_{2}O(l) + 2Br^{−}(aq) → MnO_{2}(s) + OH^{−}(aq) + 2BrO_{3}^{−}(aq)

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Balance the following equation in basic conditions. Phases are optional.

SO_{3}^{2-} + Co(OH)_{2} → Co + SO_{4}^{2-}

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Balance the following equation in basic conditions. Phases are optional.

CoCl_{2} + Na_{2}O_{2} → Co(OH)_{3} + Cl- + Na^{+}

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Balance the following redox reaction by inserting the appropriate coefficients.

SO_{4}^{2-} + NH_{3} → SO_{3}^{2-} + H_{2}O + N_{2}

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Balance the following redox reaction by inserting the appropriate coefficients.

HNO_{3} + H_{2}S → NO +S + H_{2}O

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Balance the following redox reaction by inserting the appropriate coefficients.

H^{+} + CrO_{4}^{2}^{-} + NO_{2}^{-} → Cr^{3+} + H_{2}O +NO_{3}^{-}

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Balance the following equation in acidic conditions. Phases are optional.

H_{3}AsO_{3}+ I_{2} → H_{3}AsO_{4} + I^{-}

(Redox and Oxidation Reactions)

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Balance the following equation in basic conditions. Phases are optional.

CoCl_{2 }+ Na_{2}O_{2 }→ Co(OH)_{3 }+ Cl^{-} + Na^{+}

(redox and oxidation reactions)

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How many electrons are transferred in the following reaction after balancing? (The reaction is unbalanced!)

Mg_{(s) }+ Al ^{3+}_{(aq)} → Al _{(s)} + Mg^{2+}_{(aq)}

A) 1

B) 2

C) 3

D) 4

E) 6

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Balance the following redox reaction if it occurs in basic solution. What are the coefficients in front of Br_{2} and OH^{-} in the balanced reaction?

Br_{2(l)} → BrO_{3}^{-}_{(aq)} + Br^{-}_{(aq)}

A) Br_{2} = 3, OH ^{-} = 6

B) Br_{2} = 3, OH ^{- }= 3

C) Br_{2} = 2, OH ^{-} = 5

D) Br_{2} = 1, OH ^{-} = 2

E) Br_{2} = 1, OH ^{-} = 6

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When the following redox equation is balanced with smallest whole-number coefficients, the coefficient for zinc will be.

Zn_{(s) }+ ReO_{4}^{-}_{(aq) }→ Re_{(s)} + Zn^{2+}_{(aq)} (acidic solution)

A) 2

B) 7

C) 8

D) 16

E) None of these choices is correct.

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Balance the following equation:

Cu_{(s)} + H^{+}_{(aq)} + NO_{3}^{-}_{(aq)} → NO_{(g)} + H_{2}O_{(l)} + Cu^{2+}_{(aq)}

The coefficients are:

A) 1,2,1,1,1,1

B) 3,2,1,1,1,3

C) 3,4,2,2,4,3

D) 1,8,2,2,4,1

E) 3,8,2,2,4,3

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When concentrated H_{2}SO_{4 }(l) is added to I ^{-} (aq), I _{2 }(s) and H_{2}S (g) are formed. When the equation for this reaction is balanced, what is the number of electrons transferred?

a) 8

b) 2

c) 4

d) 6

e) 1

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When this oxidation-reduction equation is the balanced in acidic solution, using only whole number coefficients, what is the coefficient for S(s)?

**?** Cr_{2}O_{7}^{2- }(aq) + **?** H_{2}S (aq) **→** **?** Cr^{3+ }(aq) + **?** S (s)

(A) 4

(B) 3

(C) 2

(D) 1

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Balance this ionic equation for a redox reaction, using only whole number coefficients.

**?** MnO_{4}^{- }(*aq*) + **?** Fe^{2+ }(*aq*) + H_{3}O^{+ }(*aq*) → **?** Mn^{2+ }(*aq*) + **?** Fe^{3+ }(*aq*) + **?** H_{2}O (*l*)

What is the coefficient for Fe^{2+} in the balanced equation?

(A) 1

(B) 3

(C) 4

(D) 5

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Consider the unbalanced redox reaction:

Cr_{2}O_{7}^{2–}(aq) + Cu(s) → Cr ^{3+} (aq) + Cu ^{2+}(aq)

Balance the equation and determine the volume of a 0.850 M K_{2}Cr_{2}O_{7} solution required to completely react with 5.25 g of Cu.

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Consider the unbalanced redox reaction:

MnO_{4}^{–}(aq) + Zn(s) → Mn^{2+}(aq) + Zn^{2+}(aq)

Balance the equation and determine the volume of a 0.500 M KMnO _{4} solution required to completely react with 2.85 g of Zn.

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Balance each redox reaction occurring in basic aqueous solution.

c. NO_{2}^{–} (aq) + Al(s) → NH_{3}(g) + AIO_{2}^{–} (aq)

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Balance each redox reaction occurring in basic aqueous solution.

b. Ag(s) + CN^{–}(aq) + O_{2}(g) → Ag(CN)_{2}^{–}(aq)

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Balance each redox reaction occurring in basic aqueous solution.

a. MnO_{4}^{–} (aq) + Br ^{–}(aq) → MnO_{2}(s) + BrO_{3}^{–} (aq)

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Balance each redox reaction occurring in acidic aqueous solution.

c. NO_{3}^{–} (aq) + Sn^{2+}(aq) → Sn^{4+}(aq) + NO(g)

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Balance each redox reaction occurring in acidic aqueous solution.

b. ClO_{4}^{–}(aq) + Cl^{–}(aq) → ClO_{3}^{–}(aq) + Cl_{2}(g)

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Balance each redox reaction occurring in acidic aqueous solution.

a. I^{–}(aq) + NO_{2}^{–} (aq) → I_{2}(s) + NO(g)

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Balance each redox reaction occurring in acidic aqueous solution.

c. MnO_{4}^{–} (aq) + Al(s) → Mn ^{2+}(aq) + Al ^{3+}(aq)

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Balance each redox reaction occurring in acidic aqueous solution.

b. Mg(s) + Cr^{3+}(aq) → Mg^{2+}(aq) + Cr(s)

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Balance each redox reaction occurring in acidic aqueous solution.

a. Zn(s) + Sn^{2+}(aq) → Zn^{2+}(aq) + Sn(s)

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Balance the following redox reaction in acidic conditions, listing the half reactions for oxidation and reduction. Also list the oxidation states of atoms in reaction.

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Balance the following redox reaction in acidic conditions, listing the half reactions for oxidation and reduction. Also list the oxidation states of atoms in reaction.

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Balance the following redox reaction occurring in acidic solution. (Hint: the Br_{2 }partakes in both half reactions)

Br_{2 }(l) → BrO_{3 }(aq) + Br ^{–}(aq)

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Design a voltaic cell from the following half-reactions:

Cr_{2}O_{7}(*aq*) + 14 H^{+}(*aq*) + 6 e ^{−} → 2 Cr^{3+}(*aq*) + 7 H_{2}O(*l*) E° = 1.33

O_{2}(*g*) + 4 H^{+}(*aq*) + 4 e ^{−} → H_{2}O(*l*) E° = 1.23

In the overall reaction, how many H ^{+} ions would be shown and which side would they be on?

A. 18 on the left

B. 16 on the left

C. 10 on the left

D. 10 on the right

E. 12 on the right

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What is the coefficient of the permanganate ion (MnO _{4}^{−}) when the following equation is balanced?

MnO_{4}^{−} + Br ^{−} → Mn^{2+} + Br_{2} (acidic solution)

A) 3 B) 4 C) 2 D) 1 E) 5

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What is the coefficient of the oxidizing agent when it is balanced within a basic solution?

Cl_{2}(g) + S_{2}O_{3} ^{2-} (aq) → Cl^{-} (aq) + SO_{4} ^{2-} (aq)

a) 1 b) 2 c) 3 d) 4

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Consider the equation: Cr_{2}O_{7} ^{2 –} + H ^{+} + I ^{–} → Cr ^{3+} (aq) + H_{2}O + I _{3}^{ –}

Which coefficient would be needed to balance I ^{–}?

a) 9 b) 7 c) 5 d) 14

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Consider the reaction of 3.0 M dichromate: Cr_{2}O_{7} ^{2–} → Cr^{3+} (Acidic Medium). What would be the coefficient on water in the balanced equation?

a. 7

b. 5

c. 6

d. 4

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