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Predict whether the following metals will react with the acid indicated. If a reaction does occur, write the net ionic equation for the reaction. Assume that reactants and products are in their standard states. (a) \(\mathrm{Ag}\) in \(\mathrm{HNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{Zn}\) in \(\mathrm{HI}(\mathrm{aq}) ;\) (c) \(\mathrm{Au}\) in \(\mathrm{HNO}_{3}\) (for the couple \(\left.\mathrm{Au}^{3+} / \mathrm{Au}, E^{\circ}=1.52 \mathrm{V}\right)\).

Short Answer

Expert verified
a) No reaction occurs between Ag and HNO3. b) The net ionic reaction between Zn and HI is: \[ Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H2_{(g)} \] c) The net ionic reaction between Au and HNO3 is : \[ 4Au_{(s)} + 24H^{+}_{(aq)} + 8NO3^-_{(aq)} \rightarrow 4Au^{3+}_{(aq)} + 2NO_{(g)} + 12H2O_{(l)} \]

Step by step solution

01

Silver (Ag) in Nitric acid (HNO3)

Silver is a low reactivity metal and does not react with nitric acid. Thus, no reaction occurs.
02

Zinc (Zn) in Hydroiodic acid (HI)

Zinc is a reactive metal and reacts with hydroiodic acid (HI) as follows: \[ Zn_{(s)} + 2HI_{(aq)} \rightarrow ZnI2_{(aq)} + H2_{(g)} \]. The net ionic equation simplifies to: \[ Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H2_{(g)} \]
03

Gold (Au) in Nitric acid (HNO3)

In nitric acid, Au does not react under normal conditions. However, with the standard potential given as \(E^{\circ}=1.52 V\), Au can react with concentrated nitric acid as follows: \[ 4Au_{(s)} + 8HNO3_{(aq)} \rightarrow 4Au(NO3)3_{(aq)} + 2NO_{(g)} + 4H2O_{(l)}\]. The net ionic equation simplifies to: \[ 4Au_{(s)} + 24H^{+}_{(aq)} + 8NO3^-_{(aq)} \rightarrow 4Au^{3+}_{(aq)} + 2NO_{(g)} + 12H2O_{(l)} \]

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Key Concepts

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

Net Ionic Equations
Understanding net ionic equations is crucial when studying chemical reactions. These equations show only the chemical species that are involved in a reaction, omitting the spectator ions that do not participate. This simplifies the equation and focuses on the essence of the chemical change. For instance, when zinc reacts with hydroiodic acid, the overall chemical equation is
\[ Zn_{(s)} + 2HI_{(aq)} \rightarrow ZnI2_{(aq)} + H2_{(g)} \].
However, the net ionic equation strips it down to the reactive species only:
\[ Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H2_{(g)} \].
By writing net ionic equations, students can better visualize and understand the actual chemical reaction taking place.
Reactivity Series
The reactivity series is a list of metals arranged in order of their ability to displace other metals in a reaction, basically, it showcases a hierarchy of metal reactivity. Metals at the top, such as potassium and lithium, are more reactive and will easily engage in chemical reactions. Conversely, metals like gold and platinum are found at the bottom, indicating low reactivity. This series is key in predicting whether a metal will react with a given acid. For example, silver (Ag) is quite low on this series, which means it won't react with dilute acids like nitric acid (HNO3). On the other hand, zinc (Zn), which is higher on the reactivity series, readily reacts with acids, as seen with hydroiodic acid (HI).
Chemical Reactions in Standard States
A standard state refers to the most stable physical state of an element or compound under standard conditions (usually 1 bar pressure and a specified temperature, typically 25°C). When we consider chemical reactions in these standard states, we can accurately measure and compare the tendencies of different substances to react. For instance, gold (Au) does not react with nitric acid under normal conditions, but based on the provided standard electrode potential (\(E^{\circ}=1.52 V\)), we can infer that under certain conditions, such as high concentration, a reaction is feasible. It's important to remember that reactivity can vary significantly with temperature, pressure, and concentration.
Redox Reactions
Redox reactions are chemical processes involving the transfer of electrons between two substances. One substance gets oxidized (loses electrons) while the other gets reduced (gains electrons). In the context of metals reacting with acids, the metal is oxidized, and the hydrogen ions from the acid are reduced. As a practical example, when zinc reacts with hydroiodic acid, zinc loses electrons to form \(Zn^{2+}\) ions, thus getting oxidized, while hydrogen ions gain electrons to form hydrogen gas (\(H2\)), getting reduced. Recognizing redox reactions is essential for understanding various chemical phenomena, including corrosion, energy storage in batteries, and metabolic processes in living organisms.

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Most popular questions from this chapter

Briefly describe each of the following ideas, methods, or devices: (a) salt bridge; (b) standard hydrogen electrode (SHE); (c) cathodic protection; (d) fuel cell.

Consider the following electrochemical cell: $$ \operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(x \mathrm{M}) | \mathrm{Ag}(\mathrm{s}) $$ (a) What is \(E_{\text {cell }}^{\circ}-\) that is, the cell potential when \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M} ?\) (b) Use the Nernst equation to write an equation for \(E_{\text {cell }}\) when \(\left[\mathrm{Ag}^{+}\right]=x\) (c) Now imagine titrating \(50.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) in the cathode half-cell compartment with 0.0100 M KI. The titration reaction is $$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgI}(\mathrm{s})$$ Calculate \(\left[\mathrm{Ag}^{+}\right]\) and then \(E_{\text {cell }}\) after addition of the following volumes of \(0.0100 \mathrm{M} \mathrm{KI}:(\mathrm{i}) 0.0 \mathrm{mL} ;(\mathrm{ii}) 20.0 \mathrm{mL}\) (iii) \(49.0 \mathrm{mL} ;(\text { iv }) 50.0 \mathrm{mL} ;(\mathrm{v}) 51.0 \mathrm{mL} ;(\mathrm{vi}) 60.0 \mathrm{mL}\) (d) Use the results of part (c) to sketch the titration curve of \(E_{\text {cell }}\) versus volume of titrant.

You prepare \(1.00 \mathrm{L}\) of a buffer solution that is \(1.00 \mathrm{M}\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) and \(1.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} .\) The solution is divided in half between the two compartments of an electrolytic cell. Both electrodes used are Pt. Assume that the only electrolysis is that of water. If 1.25 A of current is passed for 212 min, what will be the \(\mathrm{pH}\) in each cell compartment at the end of the electrolysis?

Describe in words how you would calculate the standard potential of the \(\mathrm{Fe}^{2+} / \mathrm{Fe}(\mathrm{s})\) couple from those of \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+} / \mathrm{Fe}(\mathrm{s})\).

The value of \(E_{\text {cell }}^{\circ}\) for the reaction \(\mathrm{Zn}(\mathrm{s})+\) \(\mathrm{Pb}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s})\) is \(0.66 \mathrm{V} .\) This means that for the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Pb}^{2+}(0.01 \mathrm{M})\) \(\rightarrow \mathrm{Zn}^{2+}(0.10 \mathrm{M})+\mathrm{Pb}(\mathrm{s}), E_{\text {cell }}\) equals \((\mathrm{a}) 0.72 \mathrm{V}\) (b) \(0.69 \mathrm{V} ;\) (c) \(0.66 \mathrm{V} ;\) (d) \(0.63 \mathrm{V}\)

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