Question

a) Balance the reaction shown below using the smallest possible whole-number coefficients if the reaction is carried out under acidic conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add water to one side of the reaction equation in order to balance it. $Cr_2O_7^{2-}$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $NO_3^-$ (aq) $Cr_2O_7^{2-}$ (aq) + $H^+$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $H_2O$ (l) + $NO_3^-$ (aq) b) Balance the reaction shown below using the smallest possible whole number coefficients if the reaction is carried out under basic conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add water to one side of the reaction equation in order to balance it. $Tl_2O_3$ (s) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s) + $N_2$ (g) $Tl_2O_3$(s) + $OH^-$ (aq) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s)+ $N_2$(g) + $H_2O$ (l)

          a) Balance the reaction shown below using the smallest possible whole-number coefficients if the reaction is carried out under acidic
conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add
water to one side of the reaction equation in order to balance it.
$Cr_2O_7^{2-}$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $NO_3^-$ (aq)
$Cr_2O_7^{2-}$ (aq) +  $H^+$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $H_2O$ (l) + $NO_3^-$ (aq)
b) Balance the reaction shown below using the smallest possible whole number coefficients if the reaction is carried out under basic
conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add
water to one side of the reaction equation in order to balance it.
$Tl_2O_3$ (s) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s) + $N_2$ (g)
$Tl_2O_3$(s) + $OH^-$ (aq) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s)+ $N_2$(g) + $H_2O$ (l)

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a balance the reaction shown below using the smallest possible whole number coefficients if the reaction is carried out under acidic conditions make sure that every reactant and product has 20154

Added by Joseph P.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Cheryl Glor

08/11/2023

Step 1

Step 1: a) Balance the reaction in acidic conditions: $Cr_2O_7^{2-}$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $NO_3^-$ (aq) First, balance the half-reactions: Oxidation: $NO_2^- \rightarrow NO_3^-$ $NO_2^- + H_2O \rightarrow NO_3^- + 2H^+$ $NO_2^- + H_2O Show more…

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a) Balance the reaction shown below using the smallest possible whole-number coefficients if the reaction is carried out under acidic conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add water to one side of the reaction equation in order to balance it. $Cr_2O_7^{2-}$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $NO_3^-$ (aq) $Cr_2O_7^{2-}$ (aq) + $H^+$ (aq) + $NO_2^-$ (aq) $\rightarrow$ $Cr^{3+}$ (aq) + $H_2O$ (l) + $NO_3^-$ (aq) b) Balance the reaction shown below using the smallest possible whole number coefficients if the reaction is carried out under basic conditions. Make sure that every reactant and product has a coefficient in front of it (even if the coefficient is 1). You will need to add water to one side of the reaction equation in order to balance it. $Tl_2O_3$ (s) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s) + $N_2$ (g) $Tl_2O_3$(s) + $OH^-$ (aq) + $NH_2OH$ (aq) $\rightarrow$ $TlOH$ (s)+ $N_2$(g) + $H_2O$ (l)

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Transcript

00:01 We have to balance these two equations, one in an acidic environment, the other in a basic environment.

00:08 Let's start with nitrate plus tin.

00:11 First, we're going to separate out our nitrate half equation and then our tin half equation.

00:26 So now that we got those separated, let's first balance everything but oxygen and hydrogen.

00:31 In both cases, everything's balanced.

00:35 Now let's balance our oxygen.

00:37 In our nitrate one, we have three on the left and two on the right.

00:42 So we're going to add one h2o to the left -hand side.

00:47 Now we have to add another two h's to compensate for those two that we had from the water.

00:56 Now let's take a look at charges.

01:00 For the nitrate one, we have two positives plus one negative for a net positive, and then we have zero plus zero.

01:12 So we're going to have to add some electrons.

01:15 So we're going to add one electron to the left -hand side to make everything balanced.

01:25 Now let's take a look at our tin.

01:27 We have a positive two charge on the left and a positive four charge on the right.

01:32 That means we're going to have to add some electrons to the right -hand side to bring that charge down.

01:38 So we're going to add two electrons.

01:41 Now we have to multiply that nitrate one by two to balance those electrons.

01:51 So we're going to have two electrons plus four hydrogens plus two nitrates.

01:58 That's going to yield two no2 plus two h2o.

02:07 Let's throw in our tin as well.

02:11 We don't have to worry about those electrons anymore.

02:20 We can get rid of those.

02:23 So now let's rewrite our final equation.

02:26 Sn two plus plus four h plus plus two no3 minus.

02:35 That will yield two no2 plus two h2o plus tin four plus.

02:49 Now let's move on to the basic one.

02:51 First we separate our two reactions.

02:58 One with manganese in it and then the iodine.

03:05 Let's first balance that iodine one.

03:08 We're going to put it two in front of that iodine ion on the left -hand side.

03:13 Then to the manganese half reaction, we're going to add water.

03:18 But we have to add two waters because we need two oxygens...

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