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Chapter 19: Problem 91

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{MMg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Short Answer

Expert verified
The cell potential \(E^{\circ}_{cell}\) is \(2.77 V\). In the galvanic cell, the Magnesium electrode acts as the anode (oxidation happens) and the Cadmium electrode acts as the cathode (reduction happens). The direction of the electron flow is from the magnesium electrode (anode) to the cadmium electrode (cathode).

Step by step solution

01

Identify Half Reactions and Standard Reduction Potentials

The given cell consists of a Magnesium electrode in a Magnesium Nitrate solution and a Cadmium electrode in a Cadmium Nitrate solution. Thus, the half reactions would be-\n1. Magnesium: \(Mg(s) -> Mg^{2+}(aq) + 2e^-\) [\(-2.37\,V\)]\n2. Cadmium: \(Cd^{2+}(aq) + 2e^- -> Cd(s)\) [\(0.40\,V\)]\nThe numbers in parentheses are the standard potentials \(E^{\circ}\) for these reactions, negative for oxidation and positive for reduction.
02

Calculate the Cell Potential

The reaction that occurs in the cell is the sum of the two half-reactions. To get the net cell reaction, the oxidation reaction (Magnesium) is usually reversed, and both reactions are added. Remember that the net cell potential \(E^{\circ}_{cell}\) can be found by subtracting the standard potential of the half-reaction which was reversed from the potential of the half-reaction that stayed the same, in formula format:\n\[E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\]\nThe cathode is the site where reduction takes place (Cadmium) and anode is the site where oxidation happens (Magnesium), thus\n\[E^{\circ}_{cell} = 0.40V - (-2.37V) = 2.77V\]
03

Draw a Cell Diagram

To sketch the galvanic cell include the anode (where oxidation occurs, Magnesium in this case) on one side, and the cathode (where reduction happens, Cadmium in this case) on the other. A wire connecting the two represents the flow of electrons from anode to cathode. The salt bridge, that enables the flow of ions, connects the two half-cells. The direction of electron flow will be from the magnesium electrode to the cadmium electrode, as electrons are produced at the anode (magnesium) and consumed at the cathode (cadmium).

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Most popular questions from this chapter

Fluorine \(\left(\mathrm{F}_{2}\right)\) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process. (b) What is the purpose of KF? (c) Calculate the volume of \(\mathrm{F}_{2}\) (in liters) collected at \(24.0^{\circ} \mathrm{C}\) and 1.2 atm after electrolyzing the solution for \(15 \mathrm{~h}\) at a current of \(502 \mathrm{~A}\).

A spoon was silver-plated electrolytically in a \(\mathrm{AgNO}_{3}\) solution. (a) Sketch a diagram for the process. (b) If \(0.884 \mathrm{~g}\) of \(\mathrm{Ag}\) was deposited on the spoon at a constant current of \(18.5 \mathrm{~mA},\) how long (in minutes) did the electrolysis take?

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of 4.50 A.

A galvanic cell is constructed as follows. One halfcell consists of a platinum wire immersed in a solution containing \(1.0 M \mathrm{Sn}^{2+}\) and \(1.0 M \mathrm{Sn}^{4+} ;\) the other half-cell has a thallium rod immersed in a solution of \(1.0 M \mathrm{TI}^{+}\). (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (c) What is the cell voltage if the \(\mathrm{TI}^{+}\) concentration is increased tenfold? \(\left(E_{\mathrm{T} 1^{+} / \mathrm{T} 1}^{\circ}=-0.34 \mathrm{~V} .\right)\)

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to \(\mathrm{Fe}(\mathrm{II})\) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.
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