CHM152GW6 Su09 crib

Chemistry (MasteringChemistry Series)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
CHM 152 Group Work 6 Crib Create a functional Galvanic cell given the following: On the left side of the cell there is 1.00M iron(III) nitrate solution with a solid iron electrode. On the right side, there is 1.00M cobalt(II) nitrate solution with a solid cobalt electrode. Draw the cell and label the anode, cathode, metals, solutions and all parts of the cell. Write and balance the half reactions and overall cell reaction. Calculate the cell potential E o , G o and K. Indicate where all charged particles flow in the cell and what they are. Calculate the cell potential if the cell contained 0.25M iron(III) nitrate and 1.75M cobalt(II) nitrate. Finally write the short hand notation for the cell. Reduction potentials can be found in Appendix D.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: An example cell is at the bottom of the chapter 13 review. reduction half rxn: Fe 3+ (aq) + 3e J Fe(s) E o red = -0.04V oxidation half rxn: Co(s) J Co 2+ (aq) + 2e E o ox = +0.28V total rxn: 2 Fe 3+ (aq) + 3 Co(s) J 2 Fe(s) + 3 Co 2+ (aq) E o cell = + 0.24V shorthand notation: Co(s) | Co 2+ (aq) || Fe 3+ (aq) | Fe(s) Δ G o = (-6 mol e)(96500 J/Vmole)(0.24V)( kJ / 1000J) = -1.4 x 10 2 kJ 0.24V = [ (8.314 J/molK)(298K) / (6 mol e)(96500 J/Vmol e) ] lnK ln K = 56.08717 K = 2.3 x 10 24 E = 0.24V – [ (8.314 J/molK)(298K) / (6 mol e)(96500 J/Vmol e) ] ln [ (1.75) 3 / (0.25) 2 ] = 0.22V...
View Full Document

This document was uploaded on 09/27/2011.

Ask a homework question - tutors are online