CHEM 211: CHEMISTRY VOL. 1
CHEM 211: CHEMISTRY VOL. 1
8th Edition
ISBN: 9781260304510
Author: SILBERBERG
Publisher: MCG
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Chapter 21, Problem 21.14P

(a)

Interpretation Introduction

Interpretation:

The given Cr2O72(aq)+Zn(s)Zn2+(aq)+Cr3+(aq) skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all  H+ ions.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reaction contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

Cr2O72(aq)2Cr3+(aq) ReductionZn(s)Zn2+(aq)+2e Oxidation

The atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

Cr2O72(aq)2Cr3+(aq)+7H2O(l) Zn(s)Zn2+(aq)+2e OarebalancedCr2O72(aq)+14H+(aq)2Cr3+(aq)+7H2O(l) Zn(s)Zn2+(aq)+2e Harebalanced

Cr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l) Zn(s)Zn2+(aq)+2e chargesarebalanced

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Hence the above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single equation.

Cr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l) Zn(s)Zn2+(aq)+2eCr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l) 3Zn(s)3Zn2+(aq)+6eCr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l) 3Zn(s)3Zn2+(aq)+6e_Cr2O72(aq)+14H+(aq)+3Zn(s)2Cr3+(aq)+7H2O(l)+3Zn2+(aq)Balancedreaction:__Cr2O72(aq)+14H+(aq)+3Zn(s)2Cr3+(aq)+7H2O(l)+3Zn2+(aq)

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that Cr2O72 is reduced to Cr3+ and Zn(s) is oxidized to Zn2+(aq).

Hence, the oxidizing agent is Cr2O72 and the reducing agent is Zn2+(aq).

(b)

Interpretation Introduction

Interpretation:

The given Fe(OH)2(s)+MnO42(aq)MnO2(s)+Fe(OH)3(s) skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all  H+ ions.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

MnO42(aq)MnO2(s)+2H2O(l)Fe(OH)2(s)+H2O(l)Fe(OH)3(s)+H+(aq)

The atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

MnO42(aq)MnO2(s)+2H2O(l)Fe(OH)2(s)+H2O(l)Fe(OH)3(s)+H+(aq) Oarebalanced

MnO42(aq)+4H+(aq)MnO2(s)+2H2O(l)Fe(OH)2(s)+H2O(l)Fe(OH)3(s)+H+(aq) HarebalancedMnO42(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)Fe(OH)2(s)+H2O(l)Fe(OH)3(s)+H+(aq)+e chargesarebalanced

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Hence the above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single equation.

MnO42(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)3Fe(OH)2(s)+3H2O(l)3Fe(OH)3(s)+3H+(aq)+3e_MnO42(aq)+H+(aq)+3Fe(OH)2(s)+H2O(l)MnO2(s)+3Fe(OH)3(s)Givenconditionbasic:MnO42(aq)+H+(aq)+OH(aq)+3Fe(OH)2(s)+H2O(l)MnO2(s)+3Fe(OH)3(s)+OH(aq)MnO42(aq)+3Fe(OH)2(s)+2H2O(l)MnO2(s)+3Fe(OH)3(s)+OH(aq)

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

Hence, the oxidizing agent is MnO4 and the reducing agent is Fe(OH)2.

(c)

Interpretation Introduction

Interpretation:

The given Zn(s)+NO3(aq)Zn2+(aq)+N2(g) skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all  H+ ions.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

2NO3(aq)N2(g)Zn(s)Zn2+(aq)

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

2NO3(aq)N2(g)+6H2O(l)Zn(s)Zn2+(aq) Oarebalanced

2NO3(aq)+12H+(aq)N2(g)+6H2O(l)Zn(s)Zn2+(aq) Harebalanced2NO3(aq)+12H+(aq)+10eN2(g)+6H2O(l)Zn(s)Zn2+(aq)+2e chargesarebalanced

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single equation.

2NO3(aq)+12H+(aq)+10eN2(g)+6H2O(l)5Zn(s)5Zn2+(aq)+10e_2NO3(aq)+12H+(aq)+5Zn(s)N2(g)+6H2O(l)+5Zn2+(aq)

Hence, the oxidizing agent is NO3 and the reducing agent is Zn.

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Chapter 21 Solutions

CHEM 211: CHEMISTRY VOL. 1

Ch. 21.4 - Prob. 21.6AFPCh. 21.4 - Prob. 21.6BFPCh. 21.4 - Prob. 21.7AFPCh. 21.4 - Prob. 21.7BFPCh. 21.4 - Prob. 21.8AFPCh. 21.4 - Prob. 21.8BFPCh. 21.7 - The most ionic and least ionic of the common...Ch. 21.7 - Prob. 21.9BFPCh. 21.7 - Prob. 21.10AFPCh. 21.7 - Prob. 21.10BFPCh. 21.7 - Prob. 21.11AFPCh. 21.7 - Prob. 21.11BFPCh. 21.7 - In the final steps of the ETC, iron and copper...Ch. 21.7 - Prob. B21.2PCh. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.3PCh. 21 - Water is used to balance O atoms in the...Ch. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Aqua regia, a mixture of concentrated HNO3 and...Ch. 21 - Consider the following general voltaic...Ch. 21 - Why does a voltaic cell not operate unless the two...Ch. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Consider the following voltaic cell: In which...Ch. 21 - Consider the following voltaic cell: In which...Ch. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - A voltaic cell is constructed with an Fe/Fe2+...Ch. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - What does a negative indicate about a redox...Ch. 21 - Prob. 21.37PCh. 21 - In basic solution, Se2− and ions react...Ch. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Use the emf series (Appendix D) to arrange each...Ch. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Consider the following general electrolytic...Ch. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - In the electrolysis of molten NaBr: What product...Ch. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Write a balanced half-reaction for the product...Ch. 21 - Prob. 21.99PCh. 21 - Prob. 21.100PCh. 21 - Prob. 21.101PCh. 21 - Prob. 21.102PCh. 21 - Prob. 21.103PCh. 21 - Prob. 21.104PCh. 21 - Prob. 21.105PCh. 21 - Prob. 21.106PCh. 21 - Prob. 21.107PCh. 21 - Prob. 21.108PCh. 21 - Prob. 21.109PCh. 21 - Prob. 21.110PCh. 21 - Prob. 21.111PCh. 21 - Prob. 21.112PCh. 21 - Prob. 21.113PCh. 21 - Prob. 21.114PCh. 21 - Prob. 21.115PCh. 21 - Prob. 21.116PCh. 21 - Prob. 21.117PCh. 21 - Prob. 21.118PCh. 21 - Prob. 21.119PCh. 21 - Prob. 21.120PCh. 21 - To examine the effect of ion removal on cell...Ch. 21 - Prob. 21.122PCh. 21 - Prob. 21.123PCh. 21 - Prob. 21.124PCh. 21 - Prob. 21.125PCh. 21 - Prob. 21.126PCh. 21 - Commercial electrolytic cells for producing...Ch. 21 - Prob. 21.129PCh. 21 - Prob. 21.130PCh. 21 - The following reactions are used in...Ch. 21 - Prob. 21.132PCh. 21 - Prob. 21.133PCh. 21 - Prob. 21.134PCh. 21 - Prob. 21.135PCh. 21 - If the Ecell of the following cell is 0.915 V,...Ch. 21 - Prob. 21.137PCh. 21 - Prob. 21.138PCh. 21 - Prob. 21.139PCh. 21 - Prob. 21.140PCh. 21 - Prob. 21.141PCh. 21 - Prob. 21.142PCh. 21 - Prob. 21.143PCh. 21 - Prob. 21.144PCh. 21 - Prob. 21.145PCh. 21 - Prob. 21.146PCh. 21 - Prob. 21.147PCh. 21 - Both Ti and V are reactive enough to displace H2...Ch. 21 - For the reaction ∆G° = 87.8 kJ/mol Identity the...Ch. 21 - Two voltaic cells are to be joined so that one...Ch. 21 - Prob. 21.152PCh. 21 - Prob. 21.153P
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