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Chapter 10: Problem 79

Using the EMF series on page 391, decide which of the following redox reactions is spontaneous. Explain your answer. (a) \(3 \mathrm{Ag}+\mathrm{Au}^{3+} \rightarrow \mathrm{Au}+3 \mathrm{Ag}^{+}\) (b) \(\mathrm{Au}+3 \mathrm{Ag}^{+} \rightarrow 3 \mathrm{Ag}+\mathrm{Au}^{3+}\)

Short Answer

Expert verified
Reaction (a) is spontaneous with an E° value of +0.70 V, while reaction (b) is not spontaneous with an E° value of -0.70 V. This is determined by calculating the standard cell potential (E°) for each reaction using the EMF series table and the equation E° = E°(cathode) - E°(anode). A positive E° indicates a spontaneous reaction, and a negative E° indicates a non-spontaneous reaction.

Step by step solution

01

Identify the half-reactions

First, we need to identify the half-reactions involved in each redox reaction. The given redox reactions are: (a) \( 3 \mathrm{Ag}+\mathrm{Au}^{3+} \rightarrow \mathrm{Au}+3 \mathrm{Ag}^{+} \) (b) \( \mathrm{Au}+3 \mathrm{Ag}^{+} \rightarrow 3 \mathrm{Ag}+\mathrm{Au}^{3+} \) For (a), the half-reactions are: (i) Oxidation: \( \mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^- \) (ii) Reduction: \( \mathrm{Au}^{3+} + 3e^- \rightarrow \mathrm{Au} \) For (b), the half-reactions are: (i) Oxidation: \( \mathrm{Au} \rightarrow \mathrm{Au}^{3+} + 3e^- \) (ii) Reduction: \( \mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag}\)
02

Determine E° values from the EMF series

Now we need to find the standard reduction potentials for each half-reaction from the EMF series table on page 391: For (a): - For silver: E°(Ag⁺/Ag) = +0.80 V - For gold: E°(Au³⁺/Au) = +1.50 V For (b): - For gold: E°(Au³⁺/Au) = +1.50 V - For silver: E°(Ag⁺/Ag) = +0.80 V
03

Calculate E° for the reactions using E°(cathode) - E°(anode)

For each reaction, we will determine the E° by using the E° values of the half-reactions. For (a): - E° = E°(Au³⁺/Au) - E°(Ag⁺/Ag) = (+1.50 V) - (+0.80 V) = +0.70 V For (b): - E° = E°(Ag⁺/Ag) - E°(Au³⁺/Au) = (+0.80 V) - (+1.50 V) = -0.70 V
04

Determine the spontaneity

Finally, we can determine the spontaneity of the reactions based on their E° values. For (a): As E° = +0.70 V is positive, reaction (a) is spontaneous. For (b): As E° = -0.70 V is negative, reaction (b) is not spontaneous and is not favored.

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Most popular questions from this chapter

A battery is constructed from iron, Fe, and silver, \(\mathrm{Ag}\), by dipping a strip of each metal into a solution of its ions \(\left(\mathrm{Fe}^{3+}\right.\) and \(\mathrm{Ag}^{+}\), respectively). As the battery operates, the \(\mathrm{Fe}^{3+}\) concentration increases while the \(\mathrm{Ag}^{+}\) concentration decreases. (a) What is getting oxidized? (b) What is getting reduced? (c) Draw a battery similar to the one you drew for WorkPatch \(10.7\), label which way the electrons flow, and label cathode, \(+\), and anode, \(=\).

(a) Use the following results to arrange the four metals, \(\mathrm{W}, \mathrm{X}, \mathrm{Y}, \mathrm{Z}\) in the vertical column on the right from the most active (at the top) to least active (on the bottom). \(\mathrm{W}+\mathrm{X}^{+} \rightarrow \mathrm{W}^{+}+\mathrm{X}\) most active \(\mathrm{X}+\mathrm{Z}^{+} \rightarrow \mathrm{X}^{+}+\mathrm{Z}\) \(\mathrm{Y}^{+}+\mathrm{Z} \rightarrow\) no reaction \(\mathrm{X}+\mathrm{Y}^{+} \rightarrow \mathrm{X}^{+}+\mathrm{Y}\) least active (b) Based on your series in part a, circle ALL the reactions below that you would expect to occur spontaneously as written. Explain. \(\mathrm{W}^{+}+\mathrm{Y} \rightarrow \mathrm{W}+\mathrm{Y}^{+} \quad \mathrm{W}^{+}+\mathrm{Z} \rightarrow \mathrm{W}+\mathrm{Z}^{+}\)

Which method of assigning shared electrons is correct-the double-counting method of the octet rule or the oxidation-state method?

For each redox reaction, indicate which substance is the oxidizing agent and which is the reducing agent. \(\mathrm{P}_{4}+6 \mathrm{Br}_{2} \rightarrow 4 \mathrm{PBr}_{3}\)

A student claims that in his battery, the electrons flow from the positive to the negative electrode. In fact, this is not true for any battery. Explain why electrons would never flow in this direction and how the direction of electron flow tells you that the cathode is where reduction occurs.
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