Extra Credit 5
- Page ID
- 83447
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Q19-1D
a. because M can displace Ag+ (aq), its reduction potential should be less than the reduction potentials of Ag+, because smaller reduction potential means more reactive metals. M cannot displace Cu2+(aq), its reduction potential is greater than Cu2+.
Thus, electrode potentials of Cu2+ is 0.340v and Ag is 0.8v.
0.340V <Eo<0.80V
b. cannot produce H2 so Eo< H2 and Eo>Fe. so:
-0.440 < Eo < 0
Q19-31B
We first find ΔG∘ of the reaction with the given values for standard Gibbs Free energy of each compound in the cell's reaction. Plugging in these values we get:
ΔG∘=ΔG of [PbO(s)]+2ΔG [Ag(s)]−ΔG[Pb(s)]−ΔG[Ag2O(s)]
ΔG of PbO(s)=−187.9kJ/mol+2(0)−0−(−11.20kJ/mol)=−176.7kJ/mol
Now that we have our value for the standard Gibbs free energy of the reaction, we can use the related equation below to calculate the Eocell value.
Q20-15A
Q21-23B
Q24-34A
Q25-11D
c. Cf25298+B105→262103Lr
A: 252+10 =262
Z: 98+5=103
Q25-11E
a.74Be+11H → 95B+y
b. 1712Mg+31H → 1913Al+10n
c. 2215P+10n → 2214Si+11H
Q25-42e
a. 4422Ti because it has 1:1 proton; neutron ratio and associated with Z<20
b. 3417Cl z<20 and satisfy the proton neutron ratio rule
c. 6632Ge because it has even number of neutrons, and is therefore more stable.