CLAS – Chem 1B – Chapter 11
1. Consider the following reaction: 2 VO2+ + 4 H+ + Cd → 2 VO2+ + 2 H2O + Cd2+
a. What substance is getting reduced?V in VO2+
b. How many moles of electrons are transferred?2 mole e–
2. Which is the strongest oxidizing agent?
a. Mn2+ b. Br- c. Br2 d. Ag+
3. Which is the strongest reducing agent?
a. Na+ b. Al c. Zn2+ d. F- e. Mn
4. True or false:
a. Galvanic cells spontaneously produce a current under standard conditionstrue
b. Electrons flow from the anode to cathode in a voltaic celltrue
c. Oxidation occurs at the cathodefalse
d. Corrosion of a metal occurs at the anodetrue
5. You designed a galvanic cell with silver and gold electrodes. Assign the electrodes, write the overall reaction, and calculate the standard cell potential. As the cell operates what happens to the masses of the silver and gold electrodes?
Cathode ⇒Au3+ + 3e– → AuE° (red) = 1.50 V
Anode ⇒ Ag → Ag+ + e– E° (ox) = - 0.8 V
Overall ⇒Au3+ + 3Ag → Au + 3 Ag+E° = 0.7 V
as the reaction goes forward the concentration or mass of the reactants ↓ over time – whereas the concentration or mass of the products ↑ over time
6. Using reduction potentials answer the following:
a. Is Cl2 able to reduce Cr3+? No Cl2 and Cr3+ are both oxidizing agents
b. Is Pb2+ able to oxidize Ni? Yes – E° (red) for Pb2+ = -0.13 V and the E° (ox) for Ni = 0.23 V
c. Will Au dissolve in an acidic solution? No - E° (ox) for Au = -1.5 V and theE° (red) for H+ = 0 V
d. Will Zn dissolve in an acidic solution? Yes - E° (ox) for Zn = 0.76 V and theE° (red) for H+ = 0 V
e. What can oxidize Al but not Zn?Mn2+
f. What can reduce Br2but not I2?Ag or Fe2+
7. Consider the following reaction for a voltaic cell at 75 °C.
Pb2+(aq) + 2 Cr2+(aq) Pb(s) + 2 Cr3+(aq)
a. Calculate the initial cell voltage for the above reaction if the initial concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M and [Cr3+] = 0.005 M.
Eredº = -0.13V and Eoxº = +0.5 V ⇒Erxnº = 0.37V
E = E° -
Q = = = 0.0025
E = 0.37 V - = 0.46 V
b. Will the initial potential increase, decrease or remain the same if the volume of electrodes is doubled by adding water?increasing the volume will decrease the concentrations causing Q to increase causing E to decrease
c. Will the initial potential increase, decrease or remain the same if the solid lead is cut in half?the mass of a solid has no impact on the potential
d. Will the initial potential increase, decrease or remain the same if you add NaCl causing PbCl2 to precipitate out of solution? precipitating out PbCl2 will cause a decrease in the concentration of Pb2+which causes an increase in Q causing E to decrease
8. Consider the following cell:
Cu (s) | Cu2+(aq)(0.001 M) || Fe 3+(aq)(0.02M) | Fe2+(aq) (? M), Pt(s)
Determine the concentration of the Fe2+ if the measured cell voltage at 27°C is 0.5 V.
anode: ⇒ Cu → Cu2+ + 2e––0.34 V
cathode: ⇒ Fe 3+ → Fe2+ + e– 0.77 V
overall: ⇒ Cu + 2 Fe 3+ → Cu2+ + 2 Fe2+0.43 V
E = E° - ⇒ 0.5 V = 0.43 V -
⇒= 0.042 M
9. Consider the following cell: Al(s) | Al3+ (1.0 M) | | Pb2+(1.0 M) | Pb(s)
Calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.6 M at 25 °C.
Anode: Al → Al3+ + 3e-E°ox = +1.66V
Cathode: Pb2+ + 2e-→Pb E°red = -0.13V
Cell: 2Al + 3Pb2+→ 2Al3+ + 3Pb E°cell = 1.53V
1.0M 1.0M
-3/2(0.6) +0.6
0.1M 1.6M
E = E° - ⇒
E = 1.53V - ln = 1.5 V
10. Consider the Galvanic cell at 25 °C described as follows:
X | X2+ || Y3+ | Y
Where X and Y are unknown metals. Given the standard reduction potential for Y3+ is 1.5 V and that K for the overall reaction in this Galvanic cell is 1.2 x 1020 what is the standard reduction potential of X2+?
Anode: X → X2+ + 2e-E°ox = - E°red
Cathode: Y3+ + 3e-→ Y E°red = 1.5 V
Cell: 3 X + 2 Y3+→ 3 X2+ + 2 Y E°cell = - E°red + 1.5 V
E°cell = = = - E°red + 1.5 V
E°red = 1.3 V
11. Consider two electrodes connected by a wire. One side has 0.0001 M Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).
a. Assign the electrodes
This is a concentration cell ⇒ the voltage is strictly due to the difference in concentration ⇒ at equilibrium the concentrations are equivalent ⇒ the side with lower concentration will have to increase and vice versa until the concentrations meet in the middle and there will no longer be a voltage aka equilibrium ⇒ so in order to increase the concentration on the lower side Fe needs to get oxidized making it the anode ⇒ to decrease the concentration on the other side the Fe2+ needs to be reduced making it the cathode
b. Calculate the cell voltage at 25 °C.
Anode: Fe → Fe2+ + 2e-E°ox = -0.44V
Cathode: Fe2+ + 2e-→ Fe E°red = +0.44V
Cell: Fe(anode) + Fe2+(cathode) → Fe(cathode) + Fe2+ (anode) E°cell = 0
E = E° - ⇒ E = 0 - ⇒ E = 0.15 V
12. What mass of Co forms from a solution of Co2+ when a current of 15 amps is applied for 1.15 hours?
Co2+ + 2e-→ Co
ne- = = = 0.644 mol e-
(0.644 mol e-)(1mol Co/2mole-)(58.93g Co/mol)= 19 g Co
13. How long will it take (in min) to plate out 10.0 g of Bi from a solution of BiO+ using a current of 25.0 A?
BiO+ + 2H+ + 3e-→ Bi + H2O
ne- =
10.0 g Bi xx = ⇒ t = 554 s or 9.23 min
14. It takes 24 min to plate out 9.8 g of an unknown metal (M) from a solution of MCl3when a current of 10 amps is applied. Identify the metal.
Molar mass (g/mol) can be used to identify the metal – you already have grams so you need to find the moles
ne- = = = 0.15 mol e-
0.15 mol e-x= 0.05mol M
molar mass = = 197 g/mol⇒ Au
15. What volume of gas at STP is produced from the electrolysis of water by a current of 3.5 amps in 15 minutes?
2 H2O (l) 2 H2(g) + O2(g)
ne- = = ⇒ ne- = 0.0326 mol e-
(0.0326 mol e-)( )(22.4 L/mol) = 0.55 L