Chem 160
Suggested Problems for quiz on 21 Feb
18.1 Describe a galvanic cell that uses the reaction
2 Ag+(aq) + Ni(s) 
2 Ag(s) + Ni2+(aq)
Identify the anode and the cathode half-reactions, and sketch the experimental
setup. Label the anode and cathode, indicate the direction of electron and ion
flow, and identify the sign of each electrode.
18.5 The standard cell potential at 25oC is 0.92V for the reaction
Al(s) + Cr3+(aq)
Al3+(aq) + Cr(s)
What is the standard free energy change for this reaction at 25oC?
18.6 The standard potential for the following galvanic cell is 0.92 V:
Al(s)|Al3+(aq)||Cr3+(aq)|Cr(s)
Look up the standard reduction potential for the Al3+/Al half-cell in Table
18.1, and calculate the standard reduction potential for the Cr3+/Cr half-cell.
18.49 The standard cell potential for a lead storage battery is 1.924 V. Calculate
Go (in kJ) for the cell reaction
Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq)  2 PbSO4(s) + 2 H2O(l)
18.54 Arrange the following oxidizing agents in order of increasing strength
under standard state conditions: Br2(l), MnO4-(aq), Sn4+(aq).
18.63 What reaction can occur, if any, when the following experiments are
carried out under standard-state conditions?
a. A strip of zinc is dipped into an aqueous solution of Pb(NO 3)2.
b. An acidic solution of FeSO4 is exposed to oxygen.
c. A silver wire is immersed in an aqueous solution of NiCl2.
d. Hydrogen gas is bubbled through aqueous Cd(NO3)2.
18.1 While I can’t make a nice sketch in Word, I hope you would be able to draw
something resembling the figure on p. 768. The anode is where oxidation occurs, and
Ni is undergoing the oxidation reaction. Ni is therefore the anode. By default, then,
Ag is the cathode. Electrons flow through the wire from the anode to the cathode – in
this case, from Ni to Ag+. The reaction: 2 Ag+(aq) + Ni(s)  2 Ag(s) + Ni2+(aq)
18.5 Go = -nFEo
= - (3 mol e-)(96,500 C/mol e-)(0.92 V)(1 J/1 C*V) = -266,340 J = -266 kJ
18.6 As written, indicates that Al is the anode and Cr is the cathode. The
standard reduction potential for Al3+ is -1.66. However, in this reaction,
since Al is the anode, it is being oxidized, not reduced – and when
calculating overall cell potential, we will have to reverse the sign on the
half-reaction, so the contribution of this electrode to the overall reaction
will be +1.66 V.
Now… Ecell = Eox + Ered
but we’re solving for Ered.
Ered = Ecell – Eox = 0.92 – 1.66 = -0.74 V
(this cell has a negative E value overall and therefore is not
spontaneous under standard conditions!!)
18.49 2 mol of e- are transferred during this reaction… (half reactions are
broken down on p. 786 of the text)
anode: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2 e- 0.296 V
cathode: PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e-  PbSO4(s) + 2 H2O(l)
1.628 V
TOTAL Eocell = 1.924 V (the only way you could figure this out would
be to look through the examples in the book… just wanted to use it
for practice of the equation)
G = -nFE = -(2 mol e-)(96,500C/mol e-)(1.924 V)(1 J/1 C*V)



J = -371 kJ
18.54 The higher the Eo value for a substance in the table, the better an
oxidizing agent it is (the more likely the reaction is to occur as written – it
has a higher potential of being reduced). So, higher up in the list, better
oxidizing agent. Thus, ranking our oxidizing agents:
MnO4-(aq) > Br2(l) > Sn4+(aq)
18.63 a. The two metals are Zn and Pb2+ -- see if the overall Ecell would be
positive.
Zn(s)  Zn2+(aq) + 2 e-
Eo = 0.76 V
Pb2+(aq) + 2 e-  Pb(s)
Eo = -0.13 V
TOTAL Zn(s) + Pb2+(aq)  Zn2+(aq) + Pb(s) Eocell = 0.63
YES – this would be spontaneous.
b. 4 Fe2+(aq)  4 Fe3+(aq) + 4 e-
Eo = -0.77 V
O2(g) + 4 H+(aq) + 4 e-  2 H2O(l)
Eo = 1.23 V
TOTAL
4 Fe2+(aq) + O2(g) + 4 H+(aq)  4 Fe3+(aq) + 2 H2O(l)
Eocell = 0.46 V
YES – this would be spontaneous.
c. 2 Ag(s)  2 Ag+(aq) + 2 e-
Eo = -0.80 V
Ni2(aq)+ + 2 e-  Ni(s)
Eo = -0.26 V
TOTAL
2 Ag(s) + Ni2+(aq)  2 Ag+(aq) + Ni(s) Eocell = -1.06 V
NO – negative Eocell so this is not spontaneous.
d. H2(g)  2 H+(aq) + 2 e-
Eo = 0.00 V
Cd2+(aq) + 2 e-  Cd(s)
Eo = -0.40 V
TOTAL
H2(g) + Cd2+(aq)  2 H+(aq) + Cd(s)
NO – negative Eocell so this is not spontaneous.
Eocell = -0.40 V