Science, Tech, Math Science Galvanic Cell Example Problem Constructing Galvanic Cells using Standard Reduction Potentials Share Flipboard Email Print A galvanic cell is one type of battery. ilbusca / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry in Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate by Todd Helmenstine Updated March 08, 2018 Galvanic cells are electrochemical cells which use the transfer of electrons in redox reactions to supply an electric current. This example problem illustrates how to form a galvanic cell from two reduction reactions and calculate the cell EMF.Galvanic Cell ProblemGiven the following reduction half-reactions:O2 + 4 H+ + 4 e- → 2 H2ONi2+ + 2 e- → NiConstruct a galvanic cell using these reactions. Find:a) Which half-reaction is the cathode. b) Which half-reaction is the anode.c) Write and balance the total cell redox reaction.d) Calculate E0cell of the galvanic cell.How to Find the SolutionTo be galvanic, the electrochemical cell must have a total E0cell > 0.From the Table of Common Standard Reduction Potentials:O2 + 4 H+ + 4 e- → 2 H2O E0 = 1.229 VNi2+ + 2 e- → Ni E0 = -0.257 VTo construct a cell, one of the half-reactions must be an oxidation reaction. To make a reduction half-reaction into an oxidation half-reaction, the half-reaction is reversed. The cell will be galvanic if the nickel half-reaction is reversed.E0Oxidation = - E0ReductionE0Oxidation = -(-0.257 V) = 0.257 VCell EMF = E0cell = E0Reduction + E0OxidationE0cell = 1.229 V + 0.257 VE0cell = 1.486 V**Note: If the oxygen reaction had been reversed, the E0cell would not have been positive and the cell would not be galvanic.** In galvanic cells, the cathode is the location of the reduction half-reaction and the anode is where the oxidation half-reaction takes place. Cathode: O2 + 4 H+ + 4 e- → 2 H2OAnode: Ni → Ni2+ + 2 e-To find the total reaction, the two half-reactions must be combined. O2 + 4 H+ + 4 e- → 2 H2O+ Ni → Ni2+ + 2 e-To balance the total number of electrons on both sides, the nickel half-reaction must be doubled. O2 + 4 H+ + 4 e- → 2 H2O+ 2 Ni → 2 Ni2+ + 4 e-Combine the reactions:O2(g) + 4 H+(aq) + 2 Ni(s) → 2 H2(ℓ) + 2 Ni2+(aq)Answers:a. The half-reaction O2 + 4 H+ + 4 e- → 2 H2O is the cathode.b. The half-reaction Ni → Ni2+ + 2 e- is the anode.c. The balanced cell reaction is:O2(g) + 4 H+(aq) + 2 Ni(s) → 2 H2(ℓ) + 2 Ni2+(aq)d. The cell EMF is 1.486 volts. citecite this article Format mla apa chicago Your Citation Helmenstine, Todd. "Galvanic Cell Example Problem." ThoughtCo, Mar. 8, 2018, thoughtco.com/galvanic-cell-example-problem-609494. Helmenstine, Todd. (2018, March 8). Galvanic Cell Example Problem. Retrieved from https://www.thoughtco.com/galvanic-cell-example-problem-609494 Helmenstine, Todd. "Galvanic Cell Example Problem." ThoughtCo. https://www.thoughtco.com/galvanic-cell-example-problem-609494 (accessed April 25, 2018). copy citation Continue Reading