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Chapter 5: Problem 43

Write a balanced equation for these redox reactions. (a) The oxidation of nitrite ion to nitrate ion by permanganate ion, \(\mathrm{MnO}_{4}^{-}\), in acidic solution \(\left(\mathrm{MnO}_{4}^{-}\right.\) ion is reduced to \(\mathrm{Mn}^{2+}\) ). (b) The reaction of manganese(II) ion and permanganate ion in basic solution to form solid manganese dioxide. (c) The oxidation of ethanol by dichromate ion in acidic solution, producing chromium(III) ion, acetaldehyde \(\left(\mathrm{CH}_{3} \mathrm{CHO}\right),\) and water as products.

Short Answer

Expert verified
The balanced equations for the redox reactions are: (a) \(6NO_{2}^- + 2MnO_{4}^- + 8H^+ \rightarrow 6NO_{3}^- + 2Mn^{2+} + 4H_2O \), (b) \(3Mn^{2+} + MnO_{4}^- + 2H_2O \rightarrow 4MnO_2 + 4OH^{-}\), (c) \(C_2H_5OH + Cr_{2}O_{7}^{2-} + 16H^+ \rightarrow CH_{3}CHO + 2Cr^{3+} + 11H_2O\).

Step by step solution

01

Balancing Redox Equation for Reaction (a)

Identify each half-reaction: oxidation half-reaction: nitrite ion (\(NO_{2}^-\)) to nitrate ion (\(NO_{3}^-\)); reduction half-reaction: permanganate ion (\(MnO_{4}^-\)) to manganese ion (\(Mn^{2+}\)). Balance the elements and then the charge in each half-reaction. Add the balanced half-reactions together to get the final balanced equation: \(6NO_{2}^- + 2MnO_{4}^- + 8H^+ \rightarrow 6NO_{3}^- + 2Mn^{2+} + 4H_2O \)
02

Balancing Redox Equation for Reaction (b)

Identify each half-reaction: oxidation half-reaction: manganese(II) ion (\(Mn^{2+}\)) to manganese dioxide (\(MnO_{2}\)); reduction half-reaction: permanganate ion (\(MnO_{4}^-\)) to manganese dioxide (\(MnO_{2}\)). Balance the elements and then the charge in each half-reaction. Add the balanced half-reactions together to get the final balanced equation: \(3Mn^{2+} + MnO_{4}^- + 2H_2O \rightarrow 4MnO_2 + 4OH^{-}\)
03

Balancing Redox Equation for Reaction (c)

Identify each half-reaction: oxidation half-reaction: ethanol (\(C_2H_5OH\)) to acetaldehyde (\(CH_{3}CHO\)); reduction half-reaction: dichromate ion (\(Cr_{2}O_{7}^{2-}\)) to chromium(III) ion (\(Cr^{3+}\)). Balance the elements and then the charge in each half-reaction. Add the balanced half-reactions together to get the final balanced equation: \(C_2H_5OH + Cr_{2}O_{7}^{2-} + 16H^+ \rightarrow CH_{3}CHO + 2Cr^{3+} + 11H_2O\)

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Most popular questions from this chapter

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Complete and balance these half-equations. (a) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{CO}_{2}\) (acidic solution) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\) (acidic solution) (c) \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{MnO}_{2}\) (basic solution) Indicate whether oxidation or reduction is involved.

The electrolyte in a lead storage battery must have a concentration between 4.8 and \(5.3 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if the battery is to be most effective. A \(5.00 \mathrm{mL}\) sample of a battery acid requires \(49.74 \mathrm{mL}\) of \(0.935 \mathrm{M} \mathrm{NaOH}\) for its complete reaction (neutralization). Does the concentration of the battery acid fall within the desired range? [Hint: Keep in mind that the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces two \(\mathrm{H}^{+}\) ions per formula unit.]

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)
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