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Notes taken during lecture

Course Code
Kris Quinlan

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Slide No. Notes
21 -in the chart, all are reduction half-reactions
-when E is greater than zero, reaction is spontaneous
-above the highlighted line, favoured reactions
-Note that Li+ is not a reducing agent, it is a poor oxidizing agent
-Li is the reducing agent
22 -For given reaction, the lower one will occur more readily
-Electrons will cancel out
-Final overall reaction is spontaneous
23 -the one that must be switched is the more negative one
-thus the Zn is switched, as oxidation half-cell
-Overall cell potential = 1.56 V (sum of oxidation and reduction
24 -n = number of moles electrons transferred in reaction
-CHANGE: F is PER MOL of electrons
-Postive cell potential, negative free energy value
-Units cancel so left with only joules
-Same thing for standard free energy (the equation with the degree
-Ans1: B
-Ans2: A
-Remember, opposite signs
-Sn2+ +2e -> Sn
-Cu2+ + e -> Cu+
-Ecell = 0.15 V [Cu half-reaction] + 0.14 V [Sn switched]
-No need to worry about multiplying by mole ratio because intensive
-Delta G = - nJE
-Expect answer to be negative because spontaneous reaction
-Delta G = -(2mol)(96500C/mol e)(0.29V) = -56kJ
-T can be removed in second Nernst because room temperature
25 -cannot use the standard equation because not in 1M
-Ni half reaction is the one that is flipped because it was more
-Multiply the Ag half-reaction by two but E unchanged
-Ecell = 0.25 V + 0.80 V = 1.05 V
-Q = [Ni2+]/[Ag+]squared = 11.11
-E = Enot – RT/nJ times InQ = 1.05 V – (8.314J/molK)(298K)/(2)
(98600J/mol)InQ = 1.02V
-Spontaneous in forward direction
-Increase [Ag+], shift to products, increase E cell
-Increase [Ni2+], shift to reactants, decrease E cell
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