# Practice Problem1

```Practice Problems (answer key)
A voltaic cell is constructed with Cd/Cd2+ at one half cell and Ag/Ag+ at the other. Both half
cells are operating at standard conditions.
1) Write the half reaction that takes place at the cathode.
Reduction (gaining electrons) always occurs at the cathode. The species that is reduced at
the cathode always has the most positive/ least negative standard reduction potential.
Ag+ + e- Ag(s) Eo red= +0.80 V
Cd 2+ + 2e-  Cd(s) Eo red= -0.40V
Ag + + e-  Ag(s) has the most positive/least negative standard reduction potential, so it
is reduced at the cathode.
The half reaction that takes place at the cathode is: Ag+ + e-  Ag(s)
2) At which electrode does the oxidation occur?
The oxidation always takes place at the anode. In this case the anode is the cadmium
electrode.
3) Write the balanced net ionic equation for the spontaneous redox reaction that occurs in
this voltaic cell.
Cd (s) + 2 Ag+ (aq) Cd 2+ +2 Ag(s)
Cadmium is oxidized(loses electrons) and silver is reduced (gains electrons). The silver
ion acquired a stoichiometric coefficient of 2 to balance for charges. Solid silver then
acquires a stoichiometric coefficient of 2 to balance for atoms.
4) Calculate the standard cell potential, E, in volts for the spontaneous reaction that occurs
in this cell.
Eo cell= Eo red(cathode) – E ored (anode)
Eo cell= Eo red(cathode)+ Eo oxidation (anode)
Eo cell=0.80 V+(0.40V)
E ocell= 1.20V
The standard oxidation potential (Eo oxidation) of cadmium has the same numerical value
as its standard reduction potential, but the sign changes.
Reduction: Cd2+ + 2e-  Cd(s) E ored= -0.40V
Oxidation: Cd(s) Cd2+ +2eE ooxidation = +0.40V
When a reaction is reversed the sign for the half cell potential changes.
Use the list of standard reduction potential values to indicate which species in each set is the
strongest oxidizing agent.
1) Cd 2+ or Al 3+
Cd 2+ + 2e-  Cd(s) E red= -0.40V
Al 3+ + 3e-  Al(s) E red= -1.66V
The standard half cell reduction potential for Cd2+ is less negative, so it is more easily
reduced- making it the stronger oxidizing agent.
2) Ni2+ (aq) or Cu(s)
The stronger oxidizing agent is more easily reduced. This means that it is more likely to
gain electrons. Solid metals are not easily reduced, as they do not tend to want to gain
extra electrons to form negative ions. Thus, Ni2+ is more likely to gain electron, making it
the stronger oxidizing agent.
Use the list of standard reduction potential values to indicate which species in each set is the
strongest reducing agent.
1) Al3+ (aq) or Al(s)
The stronger reducing agent is more easily oxidized, meaning it loses electrons more
easily. Solid aluminum can lose three electrons to become the aluminum ion. Al 3+ is not
easily oxidized as it does not tend to lose additional electrons. Thus, Al(s) is more likely
to lose electron, making it the stronger reducing agent.
2) Ag(s) or Ni(s)
The stronger reducing agent is more easily oxidized. This means that it loses electrons
more easily. We can solve this problem by looking at the standard half-cell oxidation
potential values. The most positive/least negative standard half-cell oxidation potential is
the strongest reducing agent (it is more likely to be oxidized- lose electrons).
Ag(s)  Ag+ + eEo ox = -0.80 V( sign changed because reaction was reversed
Ni(s)  Ni 2+ + 2e- Eo ox = +0.25 V(sign changed because reaction was reversed
Ni(s) is the strongest reducing agent, as it loses electrons more easily.
The following spontaneous reaction takes place in an acidified voltaic cell
O2 (g)+ 2 H2 S (g)  2 S (s) + 2 H2O(l)
1)What is the half reaction that takes place at the anode?
2H2S(g)  2 S(s) + 4 H+ (aq) + 4 e2)What is the half reaction that takes place at the cathode?
O2 (g) + 4 H+(aq) + 4 e-  2 H2O(l)
3)Calculate the standard cell potential, Eo cell
Eo cell= Eo red(cathode) – Eo red(anode)
E ocell= Eo red(Cathode) + Eo oxi(anode)
E ocell= 1.23V+ (-0.40V)
Eo cell= 1.09V
4) What must the partial pressures of the reactants be in order to produce the voltage in part c ?
The partial pressures of O2 and H2S must be 1.0 atm. Under standard conditions the partial
pressures of all gases must be 1.0 atm.
5) What volume of oxygen gas is consumed by the reaction if 0.128 mol of H2S is used? ( Hint:
use the ideal gas law)
0.128 mol H2Sx (1mol O2 / 2 mol H2S)= 0.0640 molO2
PV=nRT
V=Nrt/P = (0.0640 mol O2)(0.0821 Latm/molK)(298K)/ 1.0atm
V= 1.57 L O2
6) Find the total charge that passes though the cell when 0.128 moles of H2S are consumed
0.128 mol H2Sx (4 mol e- / 2mol H2S) x ( 96500C/ 1mole-) = 2.47x104 C
```