At Which Electrode Is Aluminum Produced In A Galvanic Cell And In An Electrolytic Cell?

Voltaic cells are driven by a spontaneous nadechworld.comical reaction that produces an electric current through an outside circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they are not the only kind of electronadechworld.comical cell. The reverse reaction in each case is non-spontaneous and requires electrical energy to occur.

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The general form of the reaction can be written as:

< underset{longleftarrow ext{Non spontaneous}}{overset{ ext{Spontaneous} longrightarrow}{ ext{Reactants} ightleftharpoons ext{Products} + ext{Electrical Energy}}}>

It is possible to construct a cell that does work on a nadechworld.comical system by driving an electric current through the system. These cells are called electrolytic cells. Electrolytic cells, like galvanic cells, are composed of two half-cells–one is a reduction half-cell, the other is an oxidation half-cell. The direction of electron flow in electrolytic cells, however, may be reversed from the direction of spontaneous electron flow in galvanic cells, but the definition of both cathode and anode remain the same, where reduction takes place at the cathode and oxidation occurs at the anode. Because the directions of both half-reactions have been reversed, the sign, but not the magnitude, of the cell potential has been reversed.

Electrolytic cells are very similar to voltaic (galvanic) cells in the sense that both require a salt bridge, both have a cathode and anode side, and both have a consistent flow of electrons from the anode to the cathode. However, there are also striking differences between the two cells. The main differences are outlined below:

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Quantitative Aspects of Electrolysis

Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electrical charge Q which passes through the cell. For example, the half-equation

tells us that when 1 mol Ag+ is plated out as 1 mol Ag, 1 mol e– must be supplied from the cathode. Since the negative charge on a single electron is known to be 1.6022 × 10–19 C, we can multiply by the Avogadro constant to obtain the charge per mole of electrons. This quantity is called the Faraday Constant, symbol F:

F = 1.6022 × 10–19 C × 6.0221 × 1023 mol–1 = 9.649 × 104 C mol–1

Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge Q passing through an electrode is related to the amount of electrons ne– by

Thus F serves as a conversion factor between (n_{e^-}) and (Q).

Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment. Since a coulomb is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows:

In this equation I represents current and t represents time. If you remember that

coulomb = 1 ampere × 1 second 1 C = 1 A s

you can adjust the time units to obtain the correct result. Now that we can predict the electrode half-reactions and overall reactions in electrolysis, it is also important to be able to calculate the quantities of reactants consumed and the products produced. For these calculations we will be using the Faraday constant:

1 mol of electron = 96,485 C

charge (C) = current (C/s) x time (s)

(C/s) = 1 coulomb of charge per second = 1 ampere (A)

Simple conversion for any type of problem:

Convert any given time to seconds Take the current given (A) over the seconds, <1 c = (A)/(s)> Finally use the stoichiometry conversion of 1 mol of electron = 96,485 C (Faraday”s Constant)


1) Predict the products of electrolysis by filling in the graph:

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Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)


1). Cl- chlorine H+ hydrogen

Cl- chlorine Cu2+ copper

I- iodine H+ Hhydrogen

2) 12 mol e– is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e–/Cr). Then the Faraday constant can be used to find the quantity of charge.






Q = 1.386 mol Cr ×




= 8.024 × 105 C

3) The product of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, ne–. From the first half-equation we can then find the amount of peroxydisulfuric acid, and the second leads to nH2O2 and finally to mH2O2.

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