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For the reduction half-reaction \(\mathrm{Hg}_{2}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\) \(\longrightarrow 2 \mathrm{Hg}(1), E^{\circ}=0.797 \mathrm{V} .\) Will \(\mathrm{Hg}(\mathrm{l})\) react with and dissolve in HCl(aq)? in HNO3(aq)? Explain.

Short Answer

Expert verified
Yes, Hg(l) will react with and dissolve in both HCl(aq) and HNO3(aq).

Step by step solution

01

Identify the possible reactions

HCl(aq) can be reduced to \(Cl_2(g)\) or \(H_2(g)\). Nitric acid (HNO3) can be reduced to \(NO_2(g)\), \(NO(g)\) and \(N_2O(g)\) or other forms. The half-cell reactions of these processes along with their standard reduction potentials need to be listed in order to compare them with the given reaction.
02

Compare the standard reduction potentials

Standard reduction potentials can be found using a standard reduction potential table: \( 2H^+(aq) + 2e^- -> H_2(g); E^0= 0.0 V\), \( 2Cl^-(aq) + 2e^- -> Cl_2(g); E^0= 1.36 V\), \( 2HNO3(aq) + 2H^+ + 2e^- -> 2NO_2 + 2H2O ; E^0= 0.956 V\), \( 2HNO3(aq) + 2H^+ + e^- -> H_2O + 2NO ; E^0= 0.78 V\), \( 4HNO3(aq) + e^- -> 2H2O + 4NO_2 ; E^0= 1.29 V\). If the reduction potential of the possible reaction with HCl or HNO3 is less than the reduction potential of the given reaction (0.797 V), then Hg(l) will not react with the acid. Conversely, if the reduction potential is more positive than 0.797 V, Hg(l) will react with and dissolve in the acid.
03

Conclude

The reduction potential of the reaction with HCl (1.36V) is more positive than 0.797V, meaning Hg(l) will react with and dissolve in HCl(aq). On the contrary, all possible reduction potentials of reactions with HNO3 (0.956V, 0.78V and 1.29V) are more positive, indicating that Hg(l) will also react and dissolve in HNO3(aq).

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Key Concepts

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

Chemical Reactions
At the heart of chemistry lies the concept of chemical reactions, which are processes that involve the transformation of one set of chemical substances into another. These processes can be categorized into various types, such as synthesis, decomposition, single replacement, and double replacement reactions.

Understanding chemical reactions is crucial when predicting the behavior of substances under different conditions. For instance, whether mercury (I) will react and dissolve in hydrochloric acid (HCl) or nitric acid (HNO3) can be inferred from knowing the type of chemical reaction that could take place. In the case of mercury's interactions with these acids, a single replacement reaction might occur, where mercury would potentially replace another element in a compound.
Reduction Half-Reaction
Electrochemical reactions comprise two half-reactions: oxidation and reduction. The reduction half-reaction involves the gain of electrons. In order to understand whether a metal will react with an acid, chemists look at the standard reduction potentials.

For instance, the reduction half-reaction for mercury (I) ions, \(\mathrm{Hg}_{2}^{2+} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Hg}(\ell)\), indicates that mercury ions gain electrons to become liquid mercury. The standard reduction potential associated with this half-reaction provides insight into its likelihood to proceed when compared to other potential reduction reactions, such as those involving HCl or HNO3. The potential needs to be compared to those of other potential reactants to determine which substance gets reduced preferentially.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical change. At the core of electrochemistry are the concepts of oxidation and reduction, where electrons transfer between reactants.

In the context of electrochemistry, the standard reduction potential is a measure of the tendency of a chemical species to be reduced, and it is quantified under standard conditions. A higher standard reduction potential indicates a stronger tendency to gain electrons and be reduced. Thus, by comparing the standard reduction potentials of the mercury (I) reduction half-reaction with those involving HCl and HNO3, we can predict the sequence of reactivity and determine whether mercury (I) will dissolve in those acids.

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Most popular questions from this chapter

Silver tarnish is mainly \(\mathrm{Ag}_{2} \mathrm{S}\) : $$\begin{array}{r}\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \\\E^{\circ}=-0.691 \mathrm{V}\end{array}$$ A tarnished silver spoon is placed in contact with a commercially available metallic product in a glass baking dish. Boiling water, to which some \(\mathrm{NaHCO}_{3}\) has been added, is poured into the dish, and the product and spoon are completely covered. Within a short time, the removal of tarnish from the spoon begins. (a) What metal or metals are in the product? (b) What is the probable reaction that occurs? (c) What do you suppose is the function of the \(\mathrm{NaHCO}_{3} ?\) (d) An advertisement for the product appears to make two claims: (1) No chemicals are involved, and (2) the product will never need to be replaced. How valid are these claims? Explain.

In each of the following examples, sketch a voltaic cell that uses the given reaction. Label the anode and cathode; indicate the direction of electron flow; write a balanced equation for the cell reaction; and calculate \(E_{\mathrm{cell}}^{\circ}\). (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Pb}^{2+}(\mathrm{aq})\) is displaced from solution by \(\mathrm{Al}(\mathrm{s})\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\) \(\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g})\)

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 \(\left.\mathrm{Au}^{3+} / \mathrm{Au}, E^{\circ}=1.52 \mathrm{V}\right)\).

For the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g}),\) describe the voltaic cell in which it occurs, label the anode and cathode,use a table of standard electrode potentials to evaluate \(E_{\text {cell }}^{\circ},\) and balance the equation for the cell reaction.

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

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