Lecture+l - Ch 12 - Oxidation-Reduction recap oxidation...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Ch 12 - Oxidation-Reduction recap Redox reactions Exchange of electrons (or atoms) between species e reduced in reaction oxidized in reaction oxidation agent (oxidant) reducing agent (reductant) Standard reduction potentials E o red For a half reaction To balance redox reactions, balance atoms, electrons, protons/hydroxides, calculate reaction potential (E o cell = E o anode + E o cathode ) 0.03 V salt bridge >> > < ANODE CATHODE Pt(s) Ag(s) Fe 2+ /Fe 3+ Ag + Pt | Fe2+, Fe3+ || Ag+ | Ag E cell º= E Fe2+ Fe3+ º+ E Ag+ Ag º = 0.77 V + 0.80 V = 0.03 V Cell Notation Cell Potentials
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Cell Notation for electrode deposition/corrosion Question : Enter the balanced half reaction that takes place at the anode for the electrochemical cell shown. Be sure to include electrons. Mg(s) | Mg 2+ (aq) || Cu 2+ (aq) | Cu(s) Solution : The reaction at the anode is always written on the left side of the || symbol. As the reaction at the anode is oxidation, Mg will lose electrons. The equation for the half reaction is, Mg (s) Mg 2+ (aq) + 2e - Question : Enter the cell notation for an electrochemical cell that has the electrode reactions shown. If needed, use platinum or graphite as an inert electrode. Pb 2+ (aq) + 2e - Pb (s) and Cr (s) Cr 3+ (aq) + 3e - Solution : In cell notation, the anode is written on the left side of the equation Cr(s) Cr 3+ (aq) + 3 e - is an oxidation and therefore occurs at the anode. It is therefore the anode and is an oxidation reaction. Cr (s) | Cr 3+ (aq) || Pb 2+ (aq) | Pb (s) This is corrosion corrosion deposition
Background image of page 2
Cell Notation for solution phase rxs Question : Enter the cell notation for an electrochemical cell that has the electrode reactions shown. If needed, use platinum or graphite as an inert electrode. Cl 2 (g) + 2 e- Cl - (aq) Cd(s) Cd 2+ (aq) + 2 e- Solution: The reaction Cl 2 (g) + 2 e- Cl - (aq) is a reduction and will be written on the right side of the || symbol. Notice that neither of the two phases of chlorine is solid. An electrode of platinum, Pt , must be provided. At the metal the Cl 2 will be able to pick up its electrons and split into 2 Cl - (aq) anions. Putting together both sides: || Cl - (aq) | Cl 2 (g) This is corrosion, same as before | Pt Cd(s)|Cd 2+ (aq)||Cl - (aq)|Cl 2 (g)|Pt in correct cell notation Cd|Cd 2+ ||Cl - |Cl 2 |Pt
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Free Energy and Cell Voltage The cell potential of a galvanic cell relates to the free energy of the cell reaction, Δ G = -nFE, where n is the number of moles of electrons transferred in the balanced chemical reaction and F is the Faraday constant, 96,485 C/mol.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/04/2009 for the course CHEM Chem 1C taught by Professor Farmer during the Spring '09 term at UCL.

Page1 / 18

Lecture+l - Ch 12 - Oxidation-Reduction recap oxidation...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online