Question 17.10: Chromium metal can be electroplated from an aqueous solution...
Chromium metal can be electroplated from an aqueous solution of potassium dichromate. The reduction half-reaction is
{Cr_{2}O_{7}}^{2-}(aq) + 14H^+(aq) + 12e^- \longrightarrow 2Cr(s) + 7H_{2}O
A current of 6.00 A and a voltage of 4.5 V are used in the electroplating.
a How many grams of chromium can be plated if the current is run for 48 minutes?
ANALYSIS
Information given: reduction half-reaction: ({Cr_{2}O_{7}}^{2-}(aq) + 14H^+(aq) + 12e^- \longrightarrow 2Cr(s) + 7H_{2}O) current (6.00 A); voltage (4.5 V); time in s (48 × 60)
Information implied: 1 C = 1 A \cdot s; MM Cr
Asked for: mass Cr plated
STRATEGY
1. Since mass is asked for, you may assume that you have all the information to convert the amount of electricity to moles of electrons. Moles of electrons provide the bridge that connects the amount of electricity to the stoichiometry of the chemical reaction. Use the following plan:
\text{amperes}\> (A) \xrightarrow{\times \text{time}\> (s)} \text{coulomb}\> (C) \xrightarrow{1\>\text{mol}\> e^- = 9.648 \times 10^4\> C} \text{mol}\> e^-
2. Convert mol e^- to mass of Cr using the stoichiometry of the reaction.
\text{mol}\> e^- \xrightarrow{12\> \text{mol}\> e^-/2\> \text{mol}\> Cr} \text{mol}\> Cr \xrightarrow{MM}\text{ mass}\> Cr
b How long will it take to completely convert 215 mL of 1.25 M K_{2}Cr_{2}O_{7} to elemental chromium?
ANALYSIS
Information given: reduction half-reaction: ({Cr_{2}O_{7}}^{2-}(aq) + 14H^+(aq) + 12e^- \longrightarrow 2Cr(s) + 7H_{2}O) current (6.00 A); voltage (4.5 V) K_{2}Cr_{2}O_{7}: V (0.215 L); M (1.25)
Information implied: 1 C = 1 A \cdot s
Asked for: time
STRATEGY
1. Since V and M are given for K_{2}Cr_{2}O_{7} (thus for {Cr_{2}O_{7}}^{2-}), you have enough information to convert to mole e^- using the stoichiometry of the reaction.
V \times M \longrightarrow \text{mol} \>{Cr_2O_7}^{2-} \xrightarrow{12\> \text{mol}\> e^-/1\> \text{mol}\> {Cr_2O_7}^{2-} } \text{mol}\> e^-
2. Convert mol e^- to coulombs.
\text{mol}\> e^- \xrightarrow{9.648 \times 10^4 C/1\> \text{mol}\> e^-}coulombs
3. Convert coulomb to time. Recall 1 A = 1 C/s.
c How many kilowatt-hours of electrical energy are required to plate 1.00 g of chromium?
ANALYSIS
Information given: reduction half-reaction: ({Cr_{2}O_{7}}^{2-}(aq) + 14H^+(aq) + 12e^- \longrightarrow 2Cr(s) + 7H_{2}O) current (6.00 A); voltage (4.5 V) mass Cr (1.00 g)
Information implied: 1 A = 1 C/s; 1 kWh = 3.600 \times 10^{6}\> J; 1 J = 1 C \cdot V MM for Cr
Asked for: kilowatt-hours
STRATEGY
1. Convert the mass of Cr to mol e^- using stoichiometry
\text{mass}\> Cr \xrightarrow{MM} \text{mol}\> Cr \xrightarrow{2\> \text{mol}\> Cr/12\> \text{mol}\> e^-} \text{mol}\> e^-
2. Find kWh by finding the energy in joules (C \times V) and then convert to kWh (3.600 \times 10^6\> J = 1\> kWh)
\text{mol}\> e^{-} \xrightarrow{1\> \text{mol}\> e^- = 9.648 \times 10^4\> C} \text{coulomb}\> (C) \xrightarrow{V} C \cdot V \longrightarrow J \longrightarrow kWh
