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Chapter 20: Problem 39

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 \(1 M\) 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 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
a) Anode: Sn electrode, Cathode: Cu electrode b) Sn electrode loses mass, Cu electrode gains mass c) Overall cell reaction: \( \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \) d) Cell emf under standard conditions: \(E_\mathrm{cell}^\circ = 0.473\,\mathrm{V}\)

Step by step solution

01

Identify the reduction reactions

First, we should write down the reduction half-reactions for both metals: \[ \mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu} \hspace{1cm} (E^\circ_\mathrm{Cu^{2+}/Cu} = +0.337\ \mathrm{V}) \] \[ \mathrm{Sn}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Sn} \hspace{1cm} (E^\circ_\mathrm{Sn^{2+}/Sn} = -0.136\ \mathrm{V}) \]
02

Identify the anode and cathode

In an electrochemical cell, the anode is the electrode where the oxidation occurs, and the cathode is where the reduction occurs. To determine which is the anode and which is the cathode, we compare the reduction potentials of the two half-reactions. The electrode with a higher reduction potential will be the cathode, and the one with a lower reduction potential will be the anode. In this case, \(E^\circ_\mathrm{Cu^{2+}/Cu}\) is greater than \(E^\circ_\mathrm{Sn^{2+}/Sn}\), so: - Anode: \(\mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2 \mathrm{e}^-\) - Cathode: \(\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu}\)
03

Electrode mass changes

The anode is the site of oxidation, where the metal loses electrons and goes into solution as a positive ion. Therefore, the mass of the anode will decrease. The cathode is where reduction occurs and metal cations gain electrons and plate onto the electrode as a solid. Therefore, the mass of the cathode will increase. In this cell, Sn is the anode and Cu is the cathode. So, the Sn electrode loses mass, and the Cu electrode gains mass.
04

Overall cell reaction

To find the overall cell reaction, we combine the half-reactions (paying attention to the number of electrons): \[ \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \]
05

Calculate the cell emf

We can calculate the cell emf under standard conditions using the Nernst equation: \[ E_\mathrm{cell}^\circ = E_\mathrm{cathode}^\circ - E_\mathrm{anode}^\circ \] Using the reduction potential values from Step 1, we get: \[ E_\mathrm{cell}^\circ = (+0.337\,\mathrm{V}) - (-0.136\,\mathrm{V}) = 0.473\,\mathrm{V} \] Now we have answered all the questions: a) Anode: Sn electrode, Cathode: Cu electrode b) Sn electrode loses mass, Cu electrode gains mass c) Overall cell reaction: \( \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \) d) Cell emf under standard conditions: \(E_\mathrm{cell}^\circ = 0.473\,\mathrm{V}\)

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Most popular questions from this chapter

(a) How many coulombs are required to plate a layer of chromium metal \(0.25 \mathrm{~mm}\) thick on an auto bumper with a total area of \(0.32 \mathrm{~m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in \(10.0 \mathrm{~s} ?(\mathrm{c})\) If the external source has an emf of \(+6.0 \mathrm{~V}\) and the electrolytic cell is \(65 \%\) efficient, how much electrical power is expended to electroplate the bumper?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+& 4 \mathrm{Br}^{-}(a q) \\ & E_{\text {red }}^{\circ}=-0.858 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\text {red }}^{\circ}=-0.43 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(a q) \\ & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) & \\\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (a) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.

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) \(\mathrm{Ni}^{+}(a q)-\rightarrow \rightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)\) (acidic solution) (d) \(\mathrm{Cl}_{2}(a q)-\rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q)\) (basic solution)

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) Oxidizing agents can convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

A voltaic cell is constructed that uses the following halfcell reactions: $$ \begin{gathered} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{gathered} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M}\). (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 \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?
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