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Combine the half-reaction for the reduction of \(\mathrm{O}_{2}\) \(\mathrm{O}_{2}(a q)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)\) with the following oxidation half-reactions (which are based on common iron minerals) to develop complete redox reactions: a. \(2 \mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{CO}_{2}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}\) b. \(3 \mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+3 \mathrm{CO}_{2}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}\) c. \(2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}\)

Short Answer

Expert verified
Question: Combine the given reduction half-reaction with each of the three oxidation half-reactions to develop complete redox reactions. Reduction half-reaction: O鈧 + 4H鈦 + 4e鈦 鈫 2H鈧侽 Oxidation half-reactions: 1) 2FeCO鈧 + H鈧侽 鈫 Fe鈧侽鈧 + 2CO鈧 + 2H鈦 + 2e鈦 2) 3FeCO鈧 + H鈧侽 鈫 3Fe鈧僌鈧 + 3CO鈧 + 2H鈦 + 2e鈦 + 2H鈧侽 3) 2Fe鈧僌鈧 + H鈧侽 鈫 3Fe鈧侽鈧 + 2H鈦 + 2e鈦 + 2H鈧侽 Complete redox reactions: a) 2O鈧 + 6H鈦 + 2FeCO鈧 + H鈧侽 鈫 4H鈧侽 + 2Fe鈧侽鈧 + 2CO鈧 b) O鈧 + 2H鈦 + 3FeCO鈧 + H鈧侽 鈫 3Fe鈧僌鈧 + 3CO鈧 + 2H鈧侽 c) O鈧 + 2H鈦 + 2Fe鈧僌鈧 + H鈧侽 鈫 3Fe鈧侽鈧 + 2H鈧侽

Step by step solution

01

Balancing number of electrons

To balance the number of electrons, multiply the reduction half-reaction by 2 and keep the oxidation half-reaction as is. 2(O鈧 + 4H鈦 + 4e鈦 鈫 2H鈧侽) 2FeCO鈧 + H鈧侽 鈫 Fe鈧侽鈧 + 2CO鈧 + 2H鈦 + 2e鈦
02

Adding half-reactions

Add both half-reactions together. 2O鈧 + 8H鈦 + 8e鈦 + 2FeCO鈧 + H鈧侽 鈫 4H鈧侽 + 2Fe鈧侽鈧 + 2CO鈧 + 2H鈦 + 2e鈦
03

Simplification

Cancel out 2H鈦 and 2e鈦 on both sides, and simplify the equation. 2O鈧 + 6H鈦 + 2FeCO鈧 + H鈧侽 鈫 4H鈧侽 + 2Fe鈧侽鈧 + 2CO鈧 b: Combining the reduction half-reaction with the second oxidation half-reaction:
04

Balancing number of electrons

The number of electrons in both half-reactions is the same (2e鈦) so no additional balancing is required.
05

Adding half-reactions

Add both half-reactions together. O鈧 + 4H鈦 + 4e鈦 + 3FeCO鈧 + H鈧侽 鈫 3Fe鈧僌鈧 + 3CO鈧 + 2H鈦 + 2e鈦 + 2H鈧侽
06

Simplification

Cancel out 2H鈦 and 2e鈦 on both sides, and simplify the equation. O鈧 + 2H鈦 + 3FeCO鈧 + H鈧侽 鈫 3Fe鈧僌鈧 + 3CO鈧 + 2H鈧侽 c. Combining the reduction half-reaction with the third oxidation half-reaction:
07

Balancing number of electrons

The number of electrons in both half-reactions is the same (2e鈦), so no additional balancing is required.
08

Adding half-reactions

Add both half-reactions together. O鈧 + 4H鈦 + 4e鈦 + 2Fe鈧僌鈧 + H鈧侽 鈫 3Fe鈧侽鈧 + 2H鈦 + 2e鈦 + 2H鈧侽
09

Simplification

Cancel out 2H鈦 and 2e鈦 on both sides, and simplify the equation. O鈧 + 2H鈦 + 2Fe鈧僌鈧 + H鈧侽 鈫 3Fe鈧侽鈧 + 2H鈧侽

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balance Oxidation-Reduction Equations
Understanding how to balance oxidation-reduction (redox) equations is crucial for grasping how chemical reactions involve the transfer of electrons. In redox reactions, there are two key processes: oxidation, where an element loses electrons, and reduction, where an element gains electrons. To balance these equations, each electron lost in the oxidation half must be gained in the reduction half.

When we balance redox equations, we ensure that the number of electrons lost is equal to the number of electrons gained. This means adjusting coefficients to have the same number of electrons in both the oxidation and reduction half-reactions before adding them together. In the example provided, the oxygen reduction half-reaction is multiplied by 2 to match with the 2 electrons lost in each iron mineral oxidation reactions.

It's also important to balance other atoms and charges. After aligning the number of electrons, you may need to add coefficients to other reactants and products to balance the atoms, like oxygen and hydrogen, across both sides of the final equation.
Electrochemical Cell Reactions
Electrochemical cells facilitate redox reactions, where a chemical reaction generates an electrical current or an electrical current drives a chemical reaction. The reactions involved in the exercise are similar to those that occur in electrochemical cells, such as batteries or fuel cells.

In these cells, oxidation occurs at the anode, and reduction occurs at the cathode. Electrons flow from the anode to the cathode through an external circuit, which is how electrical energy is harnessed. The reduction half-reactions provided in the exercise are akin to reactions at the cathode, and the oxidation of iron minerals represents the anode reactions. By balancing and combining these half-reactions, we are essentially depicting the overall reaction taking place in an electrochemical cell.
Chemical Stoichiometry
Chemical stoichiometry is the aspect of chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It enables us to predict the amounts of reactants needed and products formed. In the context of the provided exercise, stoichiometry is applied to deduce the proportions of oxygen, water, and iron carbonate required to form iron oxide and carbon dioxide.

Stoichiometry relies on the balanced chemical equation. For instance, in the reaction 2FeCO鈧 + H鈧侽 鈫 2Fe鈧侽鈧 + 2CO鈧 + 2H鈦 + 2e鈦, we understand that two moles of iron carbonate react with one mole of water. Once the equation is fully balanced, these stoichiometric coefficients ensure that the mass and the charge are conserved throughout the reaction, reflecting the Law of Conservation of Mass.

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Most popular questions from this chapter

A solution of table salt is a good conductor of electricity, but a solution containing an equal molar concentration of table sugar is not. Why?

Explain how a mixture of anion and cation exchangers can be used to deionize water.

Toxic cyanide ions can be removed from wastewater by adding hypochlorite. \(2 \mathrm{CN}^{-}(a q)+5 \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \mathrm{N}_{2}(g)+2 \mathrm{HCO}_{3}^{-}(a q)+5 \mathrm{Cl}^{-}(a q) $$ a. If \(1.50 \times 10^{3} \mathrm{L}\) of \(0.125 M \mathrm{OCl}^{-}\) are required to remove the \(\mathrm{CN}^{-}\) in \(3.4 \times 10^{6} \mathrm{L}\) of wastewater, what is the \(\mathrm{CN}^{-}\) concentration in the water in \(\mathrm{mg} / \mathrm{L} ?\) "b. How many milliliters of \(0.575 \mathrm{M} \mathrm{Ag} \mathrm{NO}_{3}\) would you need to add to a \(50.00 \mathrm{mL}\) aliquot of the final solution (consider the volumes simply additive) to precipitate the chloride ions formed in the reaction?

What is meant by a half-reaction?

For each of the following acid-base reactions, identify the acid and the base, and then write the overall ionic and net ionic equations. a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(\operatorname{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow\) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(a q)\)

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