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Chapter 18: Problem 37

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $C r^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)$ b. $\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)$

Short Answer

Expert verified
For reaction (a): The galvanic cell consists of Cr^3+ ions being reduced to Cr2O7^2- ions at the cathode, and Cl2 gas being oxidized to Cl^- ions at the anode. The electrons flow from the anode to the cathode. The overall balanced equation is: \(2 \: Cr^{3+} + 7 \: H2O + 3 \: Cl2 \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: Cl^-\). For reaction (b): The galvanic cell consists of Cu^2+ ions being reduced to Cu solid at the cathode, and Mg solid being oxidized to Mg^2+ ions at the anode. The electrons flow from the anode to the cathode. The overall balanced equation is given: \(Cu^{2+} + Mg \rightarrow Mg^{2+} + Cu\).

Step by step solution

01

Break down the reaction into the half-reactions

We need to identify which component is being reduced and which is being oxidized. From the overall reaction, we can deduce that: - Chromium ions (Cr^3+) are being reduced to form Chromium ion-dichromate (Cr2O7^2-) - Chlorine gas (Cl2) is being oxidized to form Chloride ions (Cl^-) So, we can write the half-reactions: - Reduction: \( 2 \: Cr^{3+} + 7 \: H2O \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: e^-\) - Oxidation: \( Cl2 + 2 \: e^- \rightarrow 2 \: Cl^- \) Now, we can find the balanced overall equation.
02

Balance the overall equation

Multiply the oxidation reaction by 3 to make sure the number of electrons lost in oxidation equals the number gained in the reduction: - Reduction: \( 2 \: Cr^{3+} + 7 \: H2O \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: e^-\) - Oxidation (multiplied by 3): \( 3 \: Cl2 + 6 \: e^- \rightarrow 6 \: Cl^- \) Now we can add the two half-reactions to obtain the overall balanced equation: \[2 \: Cr^{3+} + 7 \: H2O + 3 \: Cl2 \rightarrow Cr2O7^{2-} + 14 \: H^+ + 6 \: Cl^-\]
03

Sketch the galvanic cell

Now that we have the half-reactions and overall equation, we can sketch the galvanic cell. - The cathode is where reduction occurs, so Cr^3+ ions are reduced at the cathode. - The anode is where oxidation occurs, so Cl2 gas is oxidized at the anode. - The electrons flow from the anode (oxidation) to the cathode (reduction). #For reaction (b): Cu^2+(aq)+Mg(s)↔Mg^2+(aq)+Cu(s)#
04

Break down the reaction into the half-reactions

We can deduce that: - Magnesium solid (Mg) is being oxidized to form Magnesium ions (Mg^2+) - Copper ions (Cu^2+) are being reduced to form Copper solid (Cu) So, we can write the half-reactions: - Reduction: \( Cu^{2+} + 2 \: e^- \rightarrow Cu \) - Oxidation: \( Mg \rightarrow Mg^{2+} + 2 \: e^- \) Here, the overall balanced equation is given.
05

Sketch the galvanic cell

Now that we have the half-reactions and overall equation, we can sketch the galvanic cell. - The cathode is where reduction occurs, so Cu^2+ ions are reduced at the cathode. - The anode is where oxidation occurs, so Mg solid is oxidized at the anode. - The electrons flow from the anode (oxidation) to the cathode (reduction).

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Most popular questions from this chapter

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

A solution at \(25^{\circ} \mathrm{C}\) contains $1.0 M \mathrm{Cd}^{2+}, 1.0 M \mathrm{Ag}^{+}, 1.0 \mathrm{M}\( \)\mathrm{Au}^{3+},\( and 1.0 \)\mathrm{M} \mathrm{Ni}^{2+}$ in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

The general rule for salt bridges is that anions flow to the anode and cations flow to the cathode. Explain why this is true.

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at $25^{\circ} \mathrm{C}\( is 0.195 \)\mathrm{V}\( . What is \)\left[\mathrm{Cu}^{2+}\right] ?$ (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}$ $\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}$ b. $\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}$ $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}$
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