What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?

Short Answer

Expert verified
A galvanic cell generates electrical energy from a spontaneous chemical reaction, with electrons flowing from the anode to the cathode in the external circuit. An electrolytic cell, conversely, uses electrical energy to drive a non-spontaneous chemical reaction, with electron flow from the anode to the cathode being driven by an external power source.

Step by step solution

01

Defining a Galvanic Cell

A Galvanic cell (or voltaic cell) is a device that uses a chemical reaction to generate electrical energy. It consists of two half-cells, each containing a different electrode immersed in an electrolyte. The two half-cells are connected externally through a wire and internally through a salt bridge. In a galvanic cell, the redox reaction happens spontaneously and by this reaction, chemical energy is converted into electrical energy.
02

Defining an Electrolytic Cell

An electrolytic cell is a type of electrochemical cell that uses electrical energy to drive a non-spontaneous redox reaction. The construction of this cell is similar to a galvanic cell, but the purpose is to use electrical energy from an external source to cause a chemical change, like in the process of electrolysis.
03

Identifying the Differences

Now, let's outline the differences between these two forms of cells: \n\n1. Energy Conversion: In a galvanic cell, chemical energy is converted into electrical energy. In an electrolytic cell, electrical energy is used to produce a chemical change.\n\n2. Reaction Spontaneity: The redox reaction in a galvanic cell proceeds spontaneously, while the reaction in an electrolytic cell is non-spontaneous and requires an external energy source.\n\n3. Flow of Electrons: In a galvanic cell, electrons flow from the anode to the cathode via the external circuit, while in an electrolytic cell, the direction of electron flow is from the positive electrode (anode) to the negative electrode (cathode), driven by an external power source.

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Most popular questions from this chapter

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$ \mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2} $$ (b) If a 1.00 -g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\) \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(\mathrm{P}_{\mathrm{H}_{2}}=1.0 \mathrm{~atm} ?\)

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 M \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) solution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

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?

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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