17.2 Galvanic cells (Page 3/12)

Chemistry Page 3 / 12

\begin{array}{l} \underline{\begin{array}{l} oxidation: Mg (s) ⟶ {Mg}^{2+} (a q) + {2e}^{−} \\ reduction: {2H}^{+} (a q) + {2e}^{−} ⟶ H_{2} (g) \end{array}} \\ overall: Mg (s) + {2H}^{+} (a q) ⟶ {Mg}^{2+} (a q) + H_{2} (g) \end{array}

The cell used an inert platinum wire for the cathode, so the cell notation is

Mg (s) │ {Mg}^{2+} (a q) ║ H^{+} (a q) │ H_{2} (g) │ Pt (s)

The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes , like the platinum electrode in [link] , do not participate in the oxidation-reduction reaction and are present so that current can flow through the cell. Platinum or gold generally make good inert electrodes because they are chemically unreactive.

Using cell notation

Consider a galvanic cell consisting of

2Cr (s) + {3Cu}^{2+} (a q) ⟶ {2Cr}^{3+} (a q) + 3Cu (s)

Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?

Solution

By inspection, Cr is oxidized when three electrons are lost to form Cr ³⁺ , and Cu ²⁺ is reduced as it gains two electrons to form Cu. Balancing the charge gives

\begin{array}{l} \underline{\begin{array}{l} oxidation: 2Cr (s) ⟶ {2Cr}^{3+} (a q) + {6e}^{−} \\ reduction: {3Cu}^{2+} (a q) + {6e}^{−} ⟶ 3Cu (s) \end{array}} \\ overall: 2Cr (s) + {3Cu}^{2+} (a q) ⟶ {2Cr}^{3+} (a q) + 3Cu (s) \end{array}

Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so: $Cr (s) │ {Cr}^{3+} (a q) ║ {Cu}^{2+} (a q) │ Cu (s) .$ Oxidation occurs at the anode and reduction at the cathode.

Using Cell Notation

Consider a galvanic cell consisting of

5 {Fe}^{2+} (a q) + {MnO}_{4}^{−} (a q) + {8H}^{+} (a q) ⟶ {5Fe}^{3+} (a q) + {Mn}^{2+} (a q) + {4H}_{2} O (l)

Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?

Solution

By inspection, Fe ²⁺ undergoes oxidation when one electron is lost to form Fe ³⁺ , and MnO ₄ ⁻ is reduced as it gains five electrons to form Mn ²⁺ . Balancing the charge gives

\begin{array}{l} \underline{\begin{array}{l} oxidation: {5(Fe}^{2+} (a q) ⟶ {Fe}^{3+} (a q) + e^{−}) \\ reduction: {MnO}_{4}^{−} (a q) + {8H}^{+} (a q) + {5e}^{−} ⟶ {Mn}^{2+} (a q) + {4H}_{2} O (l) \end{array}} \\ {overall: 5Fe}^{2+} (a q) + {MnO}_{4}^{−} (a q) + {8H}^{+} (a q) ⟶ {5Fe}^{3+} (a q) + {Mn}^{2+} (a q) + {4H}_{2} O (l) \end{array}

Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode. No concentrations were specified so: $Pt (s) │ {Fe}^{2+} (a q {), Fe}^{3+} (a q) ║ {MnO}_{4}^{−} (a q {), H}^{+} (a q {), Mn}^{2+} (a q) │ Pt (s) .$ Oxidation occurs at the anode and reduction at the cathode.

Check your learning

Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.

Answer:

From the information given in the problem:
$\begin{array}{l} \underline{\begin{array}{l} anode (oxidation): Zn (s) ⟶ {Zn}^{2+} (a q) + {2e}^{−} \\ cathode (reduction): {Cu}^{2+} (a q) + {2e}^{−} ⟶ Cu (s) \end{array}} \\ overall: Zn (s) + {Cu}^{2+} (a q) ⟶ {Zn}^{2+} (a q) + Cu (s) \end{array}$
Using cell notation:
$Zn (s) │ {Zn}^{2+} (a q) ║ {Cu}^{2+} (a q) │ Cu (s) .$

This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a colorless solution and is labeled “Solution of M g C l subscript 2.” The beaker on the right contains a colorless solution and is labeled “Solution of H C l.” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. At the center of the diagram, the tube is labeled “Salt bridge. Each beaker shows a metal coils submerged in the liquid. The beaker on the left has a thin grey coiled strip that is labeled “M g coil.” The beaker on the right has a black wire that is oriented horizontally and coiled up in a spring-like appearance that is labeled “Non reactive platinum wire.” A wire extends across the top of the diagram that connects the ends of the M g strip and platinum wire just above the opening of each beaker. This wire is labeled “Conducting wire.” At the center of this wire above the two beakers near the center of the diagram is a right pointing arrow with the label “e superscript negative” at its base. Bubbles appear to be rising from the coiled platinum wire in the beaker. These bubbles are labeled “Bubbles of H subscript 2.” An arrow points down and to the right from a plus sign at the upper right region of the salt bride. An arrow points down and to the left from a negative sign at the upper left region of the salt bride. A curved arrow extends from the grey plug at the left end of the salt bridge into the surrounding solution in the left beaker. The label “Oxidation M g right pointing arrow M g superscript 2 plus plus 2 e superscript negative” appears beneath the left beaker. The label “Reduction 2 H superscript plus plus 2 e superscript negative right pointing arrow H subscript 2” appears beneath the right beaker. — The oxidation of magnesium to magnesium ion occurs in the beaker on the left side in this apparatus; the reduction of hydrogen ions to hydrogen occurs in the beaker on the right. A nonreactive, or inert, platinum wire allows electrons from the left beaker to move into the right beaker. The overall reaction is: $Mg + {2H}^{+} ⟶ {Mg}^{2+} + H_{2},$ which is represented in cell notation as: $Mg (s) │ {Mg}^{2+} (a q) ║ H^{+} (a q) │ H_{2} (g) │ Pt (s) .$

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Source: OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9

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