Write balanced net ionic equations for the following reactions in basic solution: A)...
Question:
Write balanced net ionic equations for the following reactions in basic solution:
A) MnO-4(aq)+C2H5OH(aq)-->MnO2(s)+CH3CO2H(aq)
B) H2O2(aq)+Cr2O2--->7(aq)?O2(g)+Cr3+(aq)
C) Sn2+(aq)+IO-4(aq)-->Sn4+(aq)+I-(aq)
D) PbO2(s)+Cl?(aq)-->PbCl2(s)+O2(g)
Redox Reaction:
Oxidation-reduction or redox reaction is composed of two half-reactions. The oxidation half-reaction involves loss of electrons which leads to an increase in the oxidation number of a species. The reduction half-reaction involves gain of electrons which leads to a decrease in the oxidation number of a species. The species being oxidized is called the reducing agent and the species being reduced is called the oxidizing agent.
Answer and Explanation: 1
A) The half-reactions and the balanced redox reaction are shown below.
reduction half-reaction: ({eq}\rm MnO_4^-(aq)+4H_2O(l)+3e^{-} \rightarrow MnO_2(s) + 2H_2O(l)+4OH^{-}(aq) {/eq}) x 4
- The number of Mn atoms is already balanced.
- The number of O atoms is balanced by adding 2 moles of H{eq}_2 {/eq}O on the product side.
- The number of H atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the reactant side and another 4 moles of OH{eq}^{-} {/eq} on the product side.
- The charge is balanced by adding 3 moles of electrons on the reactant side.
oxidation half-reaction: ({eq}\rm C_2H_5OH(aq) + H_2O(l)+4OH^{-}(aq) \rightarrow CH_3CO_2H(aq)+4H_2O(l)+4e^{-} {/eq}) x 3
- The number of C atoms is already balanced.
- The number of O atoms is balanced by adding 1 mole of H{eq}_2 {/eq}O on the reactant side.
- The number of H atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the product side and another 4 moles of OH{eq}^{-} {/eq} on the reactant side.
- The charge is balanced by adding 4 moles of electrons on the product side.
balanced redox reaction: {eq}\mathbf{4MnO_4^-(aq)+3C_2H_5OH(aq)\rightarrow 4MnO_2(s) +3CH_3CO_2H(aq)+4OH^{-}(aq)+H_2O(l)} {/eq}
- The reduction half-reaction is multiplied by 4 and the oxidation half-reaction is multiplied by 3 to obtain equal transfer of electrons.
- The half-reactions are added and the spectator terms are eliminated.
B) The half-reactions and the balanced redox reaction are shown below.
oxidation half-reaction: ({eq}\rm H_2O_2(aq)+2OH^{-}(aq)\rightarrow O_2(g)+2H_2O(l)+2e^{-} {/eq}) x 3
- The number of O atoms is already balanced.
- The number of H atoms is balanced by adding 2 moles of H{eq}_2 {/eq}O on the product side and another 2 moles of OH{eq}^{-} {/eq} on the reactant side.
- The charge is balanced by adding 2 moles of electrons on the product side.
reduction half-reaction: {eq}\rm Cr_2O_{7}^{2-}(aq)+14H_2O(l)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_2O(l)+14OH^{-}(aq) {/eq}
- The number of Cr atoms is balanced by multiplying the number of moles of Cr{eq}^{3+} {/eq} by 2.
- The number of O atoms is balanced by adding 7 moles of H{eq}_2 {/eq}O on the product side.
- The number of H atoms is balanced by adding 14 moles of H{eq}_2 {/eq}O on the reactant side and another 14 moles of OH{eq}^{-} {/eq} on the product side.
- The charge is balanced by adding 6 moles of electrons on the reactant side.
balanced redox reaction: {eq}\mathbf{Cr_2O_{7}^{2-}(aq)+3H_2O_2(aq)+H_2O(l)\rightarrow 2Cr^{3+}(aq)+3O_2(g)+8OH^{-}(aq)} {/eq}
- The oxidation half-reaction is multiplied by 3 to have equal number of electrons in both half-reactions.
- The half-reactions are added and the spectator terms are eliminated.
C) The half-reactions and the balanced redox reaction are shown below.
oxidation half-reaction: ({eq}\rm Sn^{2+}(aq) \rightarrow Sn^{4+}(aq)+2e^{-} {/eq}) x 4
- The number of Sn atoms is already balanced.
- The charge is balanced by adding 2 moles of electrons on the product side.
reduction half-reaction: {eq}\rm IO_4^-(aq)+8H_2O(l)+8e^{-} \rightarrow I^-(aq)+4H_2O(l)+8OH^{-}(aq) {/eq}
- The number of I atoms is already balanced.
- The number of O atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the product side.
- The number of H atoms is balanced by adding 8 moles of H{eq}_2 {/eq}O on the reactant side and another 8 moles of OH{eq}^{-} {/eq} on the product side.
- The charge is balanced by adding 8 moles of electrons on the reactant side.
balanced redox reaction: {eq}\mathbf{4Sn^{2+}(aq)+IO_4^-(aq)+4H_2O(l) \rightarrow Sn^{4+}(aq) +I^-(aq)+8OH^{-}(aq)} {/eq}
- The oxidation half-reaction is multiplied by € to have equal number of electrons in both half-reactions.
- The half-reactions are added and the spectator terms are eliminated.
D) The redox reaction can be balanced by inspection.
balanced redox reaction: {eq}\mathbf{PbO_2(s)+Cl_2(g)\rightarrow PbCl_2(s)+O_2(g)} {/eq}
Learn more about this topic:
from
Chapter 10 / Lesson 13Learn about oxidizing and reducing agents, how to find them, and how to identify them in redox reactions. Learn about the different types of redox reactions and how to identify a redox reaction.