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Chapter 20: Problem 41

A 1 M solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A 1 \(\mathrm{M}\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass, and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Short Answer

Expert verified
(a) The Cu electrode serves as the cathode, and the Sn electrode serves as the anode. (b) The Cu electrode gains mass, and the Sn electrode loses mass as the cell reaction proceeds. (c) The overall cell reaction is: Sn + Cu²⁺ → Sn²⁺ + Cu (d) The emf generated by the cell under standard conditions is \(0.48\,\text{V}\).

Step by step solution

01

(Step 1: Identify the half-reactions)

First, we need to identify the two half-reactions occurring at the electrodes. From the provided solutions, we have the following half-reactions: Cu²⁺ + 2e⁻ → Cu (Reduction half-reaction) Sn²⁺ + 2e⁻ → Sn (Reduction half-reaction)
02

(Step 2: Compare the reduction potentials)

We will consult a table of reduction potentials to determine which half-reaction would occur spontaneously. Here are the standard reduction potentials at 25°C: Cu²⁺ + 2e⁻ → Cu (Reduction): \(E^0_{red} = 0.34\) V Sn²⁺ + 2e⁻ → Sn (Reduction): \(E^0_{red} = -0.14\) V Since Cu²⁺ has a higher reduction potential, it would be more favorable to undergo the reduction and therefore Cu will act as the cathode. Conversely, Sn²⁺ will undergo the oxidation process, making it the anode.
03

(Step 3: Identify the overall reaction)

To find the overall reaction, we need to combine the reduction and oxidation half-reactions. We will balance the charges and add the reactions: Oxidation half-reaction: Sn → Sn²⁺ + 2e⁻ Reduction half-reaction: Cu²⁺ + 2e⁻ → Cu Overall cell reaction: Sn + Cu²⁺ → Sn²⁺ + Cu
04

(Step 4: Find the cell's emf )

The emf generated by a cell under standard conditions can be calculated using the difference of the standard reduction potentials: \(E^0_{cell} = E^0_{red(cathode)} - E^0_{red(anode)}\) In this case, \(E^0_{cell} = 0.34\,\text{V} - (-0.14\,\text{V}) = 0.48\,\text{V}\)
05

(Step 5: Identify the electrodes that gain and lose mass)

The cathode electrode gains mass as Cu²⁺ ions are reduced to metallic Cu and deposit on it. The anode electrode loses mass as the Sn metal is oxidized to Sn²⁺ ions, leaving the metal surface. (a) The Cu electrode serves as the cathode, and the Sn electrode serves as the anode. (b) The Cu electrode gains mass, and the Sn electrode loses mass as the cell reaction proceeds. (c) The overall cell reaction is: Sn + Cu²⁺ → Sn²⁺ + Cu (d) The emf generated by the cell under standard conditions is \(0.48\,\text{V}\).

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert CO into \(\mathrm{CO}_{2}\) .

A voltaic cell utilizes the following reaction: $$ \mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s) $$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of AgNO \(_{3}\) is added to the cathode half-cell, increasing the quantity of Ag' but not changing its concentration. (d) HCl is added to the AgNO\(_{3}\) solution precipitating some of the Ag' as AgCl.

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH}),\) iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) . (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?\) (b) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is - 2 .

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{c}{\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q)}\end{array} $$ The cell is operated at 298 \(\mathrm{K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 M\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential?

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?
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