Skip to main content
Ch.20 - Electrochemistry

Chapter 20, Problem 28b

Complete and balance the following equations, and identify the oxidizing and reducing agents: As2O31s2 + NO3-1aq2 ¡ H3AsO41aq2 + N2O31aq2 (acidic solution)

Verified Solution
Video duration:
7m
This video solution was recommended by our tutors as helpful for the problem above.
429
views
Was this helpful?

Video transcript

Hi everyone here we have a question asking us to balance the following redox reaction under acidic conditions and identify the oxidizing and reducing agents. So first we need to separate the whole reaction into half reactions. So we are going to have bismuth hydroxide forming bismuth and we're going to have tin oxide forming stand eight. And our next step is to balance the non hydrogen and non oxygen elements first and those are already balanced so we can move on to the next step. So now we have to balance oxygen by adding water to the size that need oxygen. So we're going to have bismuth hydroxide forming bismuth plus three water and we're going to have 10 oxide plus water forming Stand Aid Next we need to balance our hydrogen by adding hydrogen to the side that needs it. So we'll have business hydroxide plus three hydrogen forms bismuth plus three water and we'll have tin oxide plus water Representing stay and eight plus two hydrogen. Now we need to balance the charges. Were going to add electrons to the more positive side. So we are going to have bismuth hydroxide plus three hydrogen plus three electrons forms bismuth plus three water. We're going to have 10 oxide plus water forms stand eight plus two hydrogen plus two electrons. Now we need to balance the electrons on the two half reactions. So we have three for our first reaction and we have two for our second reaction. So we need to multiply the whole thing by two and this whole thing by three. So that gives us three bismuth hydroxide plus That gives us two business bismuth hydroxide plus six hydrogen plus six electrons, forms to business plus six water. And that gives us three tin oxide plus three water forms three stand eight plus six hydrogen plus six electrons. And now we need to get the overall reaction by adding the two. So our six hydrogen are going to cancel out and our six electrons are going to cancel out. Giving us two bismuth hydroxide plus 3 ox ill forms to bismuth plus three water Plus three Stand 8. And loss of elections is oxidation. So our reducing agent is our tin oxide and gain of electrons reductions are oxidizing agent is bismuth hydroxide. And those are our final answers. Thank you for watching. Bye.
Related Practice
Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr2O72-1aq2 + I-1aq2 ¡ Cr3+1aq2 + IO3-1aq2 (acidic solution)

770
views
Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: MnO4-1aq2 + Br -1aq2 ¡ MnO21s2 + BrO3-1aq2 (basic solution)

703
views
Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents:


a. MnO4−(𝑎𝑞)+CH3OH(𝑎𝑞)⟶Mn2+(𝑎𝑞)+HCOOH(𝑎𝑞)(acidic solution)

b. As2O3(𝑠)+NO3−(𝑎𝑞)⟶H3AsO4(𝑎𝑞)+N2O3(𝑎𝑞)(acidic solution)

c. Pb(OH)42−(𝑎𝑞)+ClO−(𝑎𝑞)⟶PbO2(𝑠)+Cl−(𝑎𝑞)(basic solution)

Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.) Cr2O72 - 1aq2 + CH3OH1aq2 ¡ HCOOH1aq2 + Cr3+1aq2 (acidic solution)

554
views
Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)


a. S(s) + HNO3(aq) → H2SO3(aq) + N2O(g) (acidic solution)

b. BrO3-(aq) + N2H4(g) → Br-(aq) + N2(g) (acidic solution)

c. H2O2(aq) + ClO2(aq) → ClO20(aq) _ O2(g) (basic solution)

Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.) H2O21aq2 + ClO21aq2 ¡ ClO2-1aq2 + O21g2 (basic solution)

491
views