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Chapter 20: Problem 41

A \(1 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) is placed in a beaker with a strip of Ag metal. A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a second beaker with a strip of Cu metal. A salt bridge connects the two beakers, and wires to a voltmeter link the 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
The anode is the copper electrode and the cathode is the silver electrode. The copper electrode loses mass and the silver electrode gains mass as the reaction proceeds. The overall cell reaction is \(Cu(s) + 2 Ag^{+}(aq) -> Cu^{2+}(aq) + 2 Ag(s)\). The emf generated by the cell under standard conditions is +0.46 V.

Step by step solution

01

Determine the half-reactions for the reduction and oxidation processes

The half-reactions for the reduction and oxidation processes are represented as follows: \(Ag^{+}(aq) + e^{-} -> Ag(s)\) \(Cu^{2+}(aq) + 2 e^{-} -> Cu(s)\)
02

Identify the anode and cathode by comparing the standard reduction potentials

To determine which electrode serves as the anode and which serves as the cathode, we need to compare the standard reduction potentials. Consulting a standard reduction potential table, we find the following values: For silver: \(E^{0}_{Ag+ -> Ag} = +0.80V\) For copper: \(E^{0}_{Cu^{2+} -> Cu} = +0.34V\) Since silver has a higher standard reduction potential, it will serve as the cathode, while copper will serve as the anode.
03

Determine which electrode gains or loses mass

At the anode, oxidation occurs, and atoms of the metal will lose electrons and enter the solution as ions. This means that the anode (copper) will lose mass as the reaction proceeds. At the cathode, reduction occurs, and metal ions in the solution will gain electrons and become solid metal atoms. This means that the cathode (silver) will gain mass as the reaction proceeds.
04

Write the overall balanced cell reaction

Now we need to balance the half-reactions and write the overall cell reaction. The balanced half-reactions are: Anode (oxidation): \(Cu(s) -> Cu^{2+}(aq) + 2 e^{-}\) Cathode (reduction): \(2 Ag^{+}(aq) + 2 e^{-} -> 2 Ag(s)\) Combining the two half-reactions, we get the balanced overall cell reaction: \[ Cu(s) + 2 Ag^{+}(aq) -> Cu^{2+}(aq) + 2 Ag(s) \]
05

Calculate the emf generated by the cell under standard conditions

To calculate the emf generated by the cell under standard conditions, we can use the Nernst equation: \[E_{cell} = E^{0}_{cathode} - E^{0}_{anode}\] Using the standard reduction potentials for silver (\(+0.80 V\)) and copper (\(+0.34 V\)), we get: \[E_{cell} = (+0.80 V) - (+0.34 V) = +0.46 V\] Therefore, the emf generated by the cell under standard conditions is +0.46 V.

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Most popular questions from this chapter

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: $\mathrm{MnO}_{4}^{-}(a q), \mathrm{O}_{3}(g), \mathrm{HSO}_{4}^{-}(a q), \mathrm{O}_{2}(g), \mathrm{HClO}(a q)$ (b) Arrange the following in order of increasing strength as reducing agents in basic solution: $\mathrm{Cr}(\mathrm{OH})_{3}(s), \mathrm{Fe}(s), \mathrm{Ca}(s),\( \)\mathrm{H}_{2}(g), \mathrm{Mn}(s)$

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? (c) What process occurs at the anode in the electrolysis of molten \(\mathrm{NaCl}\) ? (d) Why is sodium metal not obtained when an aqueous solution of \(\mathrm{NaCl}\) undergoes electrolysis?

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of $E_{\text {red }}^{\circ}\( for the cathode reaction is \)+0.098 \mathrm{~V}$. The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) $\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)$ (b) $\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)$ (acidic solution) (c) $\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)$ (acidic solution) (d) $\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)$ (acidic solution)

(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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