Complete and balance the following equations, and identify...

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)$ (acidic solution) (b) $\mathrm{MnO}_{4}^{-}(a q) \pm \mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow$ $\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)$ (acidic solution) (c) $\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-{ }^{-\rightarrow} \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ (acidic solution) (d) $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q)-\cdots$ $\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)$ (acidic solution) (e) $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{BrO}_{3}^{-}(a q)$ (basic solution) (f) $\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)$ (basic solution)

Complete and balance the following equations, and identify the oxidizing and reducing agents:
(a) $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)$
(acidic solution)
(b) $\mathrm{MnO}_{4}^{-}(a q) \pm \mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow$
$\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)$ (acidic solution)
(c) $\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-{ }^{-\rightarrow} \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ (acidic solution)
(d) $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q)-\cdots$
$\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)$ (acidic solution)
(e) $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{BrO}_{3}^{-}(a q)$
(basic solution)
(f) $\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)$
(basic solution)

Key Concepts

Balancing Redox Equations

Balancing redox equations requires ensuring both mass and charge balance. This can involve adding water molecules, hydrogen ions (in acidic conditions), or hydroxide ions (in basic conditions) along with electrons, depending on the medium. Mastery of this concept is essential for accurately representing chemical reactions and understanding their stoichiometry.

The Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox reactions. It involves dividing the overall reaction into two separate half-reactions—one for oxidation and one for reduction—balancing them individually for mass and charge, and then recombining them to ensure that electrons are conserved. This method simplifies the balancing process, especially in complex reactions.

Acidic and Basic Media in Redox Reactions

The medium in which a redox reaction occurs significantly affects the balancing process. In acidic solutions, hydrogen ions (H+) and water molecules are typically used to balance hydrogen and oxygen atoms. In basic solutions, hydroxide ions (OH-) and water are employed. Recognizing the medium is crucial because it dictates the steps and species required to achieve a balanced equation.

Oxidation and Reduction

Oxidation refers to the loss of electrons, while reduction denotes the gain of electrons. These complementary processes occur simultaneously in any redox reaction and are fundamental to the study of chemical reactivity. Recognizing changes in oxidation states helps determine which species are undergoing oxidation or reduction.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species. In these reactions, one species loses electrons (oxidation) while another gains electrons (reduction). Understanding redox reactions requires tracking changes in oxidation states, which provides insight into which atoms or ions are oxidized and which are reduced.

Oxidizing and Reducing Agents

An oxidizing agent is the substance that causes oxidation by accepting electrons, and is itself reduced in the process. Conversely, a reducing agent donates electrons, causing the reduction of another species and itself undergoing oxidation. Identifying these agents is key to understanding the overall electron flow in a redox reaction.

Transcript

00:02 Okay, first part is cr 207 2 negative aqueous plus i negative aquas plus 8h positive giving to cr 3 positive aquas plus i .3 negative aquas plus 4h2o liquid oxidizing agent is cr2 7 to negative and reducing agent is i negative next reaction is mn o 4 negative aquas plus 8 h positive aquas plus 5 electron giving mn 2 positive aquas plus 4 h2 o and then the second one is c h 3 ohh plus h 2 o giving h c o 2h plus 4h positive plus 4 electrons so this will be multiplied by 5 and this one will be multiplied by 4 this one already has 5 so the overall reaction now becomes 4mno 4 negative plus 5 c h 3 ohh a quers plus 12 h positive giving 4 mn 2 positive plus 5h co2h aquas plus 11 h2o liquid.

02:29 So in this case, oxidizing agent, we'll write it as oa now on words, is mno4 negative and reducing agent is ch3oh.

02:58 Moving on to the next part, i2 solid, plus 6h2o liquid, giving 2 i .03 aqueous, plus 12h positive aquas, plus 10 electrons, plus 2 electrons, giving cl negative aquas, plus h2o liquid, giving i2 solid, plus 5o ,0, cl, negative aquas plus h2 liquid gives 2 i .03 negative aquas plus 5cl negative plus 2h positive aquas so oxidizing agent here is ocel negative and reducing agent here is i2 then d part as 203 solid plus 5h2 liquid giving 2 h3 as4 a quoth plus 4h positive aquas plus 4 electrons and this is nitrate a plus 6h positive that was plus 6 electrons sorry not 6 4 actually gives n2aquoise plus 3h2o liquid so overall it's going to be as 203 solid plus 2 no3 negative a quers plus 2h2 liquid plus 2h positive aquas giving to h3 aso4 aquas plus n203 aqueurs so oxidizing agent here is no 3 negative and reducing agent is as2 -4 3 okay so next part mno 4 negative a quers plus h2o liquid plus three electron this will be to h2 will be mn 02 solid plus 4 oh h negative and the next one is br negative equals plus 6 oh negative equals giving the r03 negative plus 3h2 plus 6 electrons so here there are 6 electrons and in the upper case there are 3 electrons we multiply this with 2 so overall you get 2m...

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