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Write the equation for the reaction, if any, that occurs when each of the following experiments is performed under standard conditions. (a) Crystals of iodine are added to an aqueous solution of potassium bromide. (b) Liquid bromine is added to an aqueous solution of sodium chloride. (c) A chromium wire is dipped into a solution of nickel(II) chloride.

Short Answer

Expert verified
Answer: I2 + 2KBr → 2KI + Br2

Step by step solution

01

(a) Iodine crystals added to the potassium bromide solution

In this experiment, the iodine (I2) is a halogen and can potentially undergo a single displacement reaction with the potassium bromide (KBr) to form potassium iodide (KI) and bromine (Br2). The balanced chemical equation for this reaction is: I2 + 2KBr → 2KI + Br2
02

(b) Liquid bromine added to the sodium chloride solution

In this experiment, the liquid bromine (Br2) can potentially undergo a single displacement reaction with the sodium chloride (NaCl) to form sodium bromide (NaBr) and chlorine (Cl2). However, since bromine is not more reactive than chlorine in the halogen series, no reaction will occur. So, no chemical equation is required for this case.
03

(c) Chromium wire dipped into a solution of nickel(II) chloride

In this case, the chromium has a potential to undergo a single displacement reaction with the nickel(II) chloride (NiCl2) if chromium has greater reactivity than nickel in the reactivity series. Since chromium is more reactive than nickel, the reaction will take place, and the balanced chemical equation for this reaction is: Cr(s) + 2NiCl2(aq) → CrCl2(aq) + 2Ni(s)

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Key Concepts

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

Single Displacement Reaction
When a more reactive element displaces a less reactive element from a compound, the process is known as a single displacement reaction. In other words, one element 'swaps places' with another in a compound, often yielding a new element and a new compound in the process. For example, if we take iodine crystals and introduce them to an aqueous solution of potassium bromide, a single displacement reaction occurs. The iodine (I2) displaces the bromine (Br-) ion, forming new compounds and releasing bromine as a diatomic molecule.

The reaction, outlined step by step, is as follows: elemental iodine reacts with the potassium bromide to create potassium iodide and elemental bromine. The chemical equation is represented as: I2 + 2KBr → 2KI + Br2.

In order for a single displacement reaction to occur, the element that is doing the displacing must be more reactive than the element that is being displaced. This is why we observe this type of reaction with iodine and potassium bromide, where iodine is more reactive than bromine in this context.
Reactivity Series
The reactivity series is an empirical arrangement of metals (and some non-metals, like halogens) based on their reactivity. This series helps us predict the outcomes of certain reactions, such as single displacement reactions, by providing a hierarchy of reactivity.

For instance, in the case of chromium dipping into a solution of nickel(II) chloride, we look at the reactivity series to determine if a reaction will occur. Chromium appears above nickel in the series, indicating chromium is more reactive and can displace nickel from its compound. Consequently, the reaction proceeds with chromium forming a new compound (chromium(II) chloride) and nickel being released as a solid: Cr(s) + 2NiCl2(aq) → CrCl2(aq) + 2Ni(s).

The reactivity series is crucial for predicting and understanding such reactions. It is important for students to be familiar with this series to correctly predict product formation in displacement reactions.
Balancing Chemical Equations
A balanced chemical equation ensures that the same number of atoms of each element is present on both sides of the equation, following the law of conservation of mass. This means, in a chemical reaction, the mass and the number of atoms are conserved from reactants to products.

When we balance the equation I2 + 2KBr → 2KI + Br2, we ensure that there are two bromine atoms on each side and the same for other elements involved. The correct coefficients—numbers in front of the chemical formulas—are critical, as they indicate the right proportions of reactants and products. Balancing is a systemic process, often starting with elements that appear in only one reactant and one product, and then proceeding to balance more complex molecules. Mastery of this skill is vital in the field of chemistry to accurately describe the stoichiometry of a reaction.
Halogen Reactivity
Halogen reactivity is an important concept, particularly when dealing with single displacement reactions involving halogens. Halogens are a specific group of nonmetals in Group 17 of the periodic table and include fluorine, chlorine, bromine, iodine, and astatine. Their reactivity decreases as we move down the group, with fluorine being the most reactive and astatine the least.

This trend explains why iodine can displace bromine in a reaction with potassium bromide but not vice versa. In contrast, when we consider the reaction in which liquid bromine is added to an aqueous solution of sodium chloride, no reaction takes place. This is because bromine is less reactive than chlorine, hence it cannot displace chlorine from its compound. Understanding the order of halogen reactivity is essential in predicting whether a particular halogen will undergo a single displacement reaction with a given halide solution.

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Most popular questions from this chapter

A solution containing a metal ion \(\left(M^{2+}(a q)\right)\) is electrolyzed by a current of \(7.8\) A. After \(15.5\) minutes, \(2.39 \mathrm{~g}\) of the metal is plated out. (a) How many coulombs are supplied by the battery? (b) What is the metal? (Assume \(100 \%\) efficiency.)

Consider a voltaic cell in which the following reaction takes place. $$ 2 \mathrm{NO}_{3}^{-}(a q)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+2 \mathrm{OH}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O} $$ (a) Calculate \(E^{\circ}\). (b) Write the Nernst equation for the cell. (c) Calculate \(E\) under the following conditions: \(\left[\mathrm{NO}_{3}^{-}\right]=0.0315 M\), \(P_{\mathrm{NO}}=0.922 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.437 \mathrm{~atm}, \mathrm{pH}=11.50 .\)

For the cell $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}\right| \mathrm{Cu}^{2+} \mid \mathrm{Cu} $$ \(E^{\circ}\) is \(1.10 \mathrm{~V}\). A student prepared the same cell in the lab at standard conditions. Her experimental \(E^{\circ}\) was \(1.0 \mathrm{~V}\). A possible explanation for the difference is that (a) a larger volume of \(\mathrm{Zn}^{2+}\) than \(\mathrm{Cu}^{2+}\) was used. (b) the zinc electrode had twice the mass of the copper electrode. (c) \(\left[\mathrm{Zn}^{2+}\right]\) was smaller than \(1 M\). (d) \(\left[\mathrm{Cu}^{2+}\right]\) was smaller than \(1 M\). (e) the copper electrode had twice the surface area of the zinc electrode.

Consider a cell in which the reaction is $$ 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) $$ (a) Calculate \(E^{\circ}\) for this cell. (b) Chloride ions are added to the \(\mathrm{Ag} \mid \mathrm{Ag}^{+}\) half- cell to precipitate \(\mathrm{AgCl}\). The measured voltage is \(+0.060 \mathrm{~V}\). Taking \(\left[\mathrm{Cu}^{2+}\right]=1.0 \mathrm{M}\), calculate \(\left[\mathrm{Ag}^{+}\right]\). (c) Taking \(\left[\mathrm{Cl}^{-}\right]\) in \((\mathrm{b})\) to be \(0.10 M\), calculate \(K_{\text {sp }}\) of \(\mathrm{AgCl}\).

In a nickel-cadmium battery (Nicad), cadmium is oxidized to \(\mathrm{Cd}(\mathrm{OH})_{2}\) at the anode, while \(\mathrm{Ni}_{2} \mathrm{O}_{3}\) is reduced to \(\mathrm{Ni}(\mathrm{OH})_{2}\) at the cathode. A portable CD player uses \(0.175\) amp of current. How many grams of \(\mathrm{Cd}\) and \(\mathrm{Ni}_{2} \mathrm{O}_{3}\) are consumed when the CD player is used for an hour and a half?

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