Chromium metal can be electroplated from an aqueous solution of potassium dichromate. The reduction half-reaction is Cr2O7 2-(aq) + 14H+(aq) + 12e- → 2Cr(s) + 7H2O A current of 6.00 A and a voltage of 4.5 V are used in the electroplating. ⓐ How many grams of chromium can be plated if the current is

Question

Chromium metal can be electroplated from an aqueous solution of potassium dichromate. The reduction half-reaction is

[latex]Cr_{2}O_{7}^{2-}(aq) + 14H^{+}(aq) + 12e^{-} → 2Cr(s) + 7H_{2}O[/latex]

A current of 6.00 A and a voltage of 4.5 V are used in the electroplating.

ⓐ How many grams of chromium can be plated if the current is run for 48 minutes?

ⓑ How long will it take to completely convert 215 mL of 1.25 M [latex]K_{2}Cr_{2}O_{7}[/latex] to elemental chromium?

ⓐ

ANALYSIS
reduction half-reaction: [latex](Cr_{2}O_{7}^{2-}(aq) + 14H^{+}(aq) + 12e^{-} → 2Cr(s) + 7H_{2}O)[/latex] current (6.00 A); voltage (4.5 V); time in s (48 × 60)	Information given:
1 C = 1 A · s; MM Cr	Information implied:
mass Cr plated	Asked for:

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:

[latex] amperes (A) \underrightarrow{× time (s)} coulomb (C) \underrightarrow{1 mol e^{-} = 9.648 × 10^{4} C} mol e^{-}[/latex]
2. Convert mol [latex]e{-}[/latex] to mass of Cr using the stoichiometry of the reaction.

[latex]mol e^{-} \underrightarrow{12 mol e^{-}/2 mol Cr} mol Cr \underrightarrow{MM} mass Cr[/latex]

ⓑ

ANALYSIS
reduction half-reaction: [latex](Cr_{2}O_{7}^{2-}(aq) + 14H^{+}(aq) + 12e^{-} → 2Cr(s) + 7H_{2}O)[/latex] current (6.00 A); voltage (4.5 V) [latex]K_{2}Cr_{2}O_{7}[/latex]: V (0.215 L); M (1.25)	Information given:
1 C = 1 A · s	Information implied:
time	Asked for:

STRATEGY

1. Since V and M are given for [latex]Cr_{2}O_{7}^{2-}[/latex] (thus for [latex]K_{2}Cr_{2}O_{7}[/latex]), you have enough information to convert to mole [latex]e^{-}[/latex] using the stoichiometry of the reaction.

V × M → mol [latex]Cr_{2}O_{7}^{2-} \underrightarrow{12 mol e^{-}/1 mol Cr_{2}O_{7}^{2-}} mol e^{-}[/latex]

2. Convert mol [latex]e^{-}[/latex] to coulombs.

[latex]mol e^{-} \underrightarrow{9.648 × 10^{4} C/1 mol e^{-}} coulombs[/latex]

3. Convert coulomb to time. Recall 1 A = 1 C/s.

ⓒ

ANALYSIS
reduction half-reaction: [latex](Cr_{2}O_{7}^{2-}(aq) + 14H^{+}(aq) + 12e^{-} → 2Cr(s) + 7H_{2}O)[/latex] current (6.00 A); voltage (4.5 V) mass Cr (1.00 g)	Information given:
1 A = 1 C/s; 1 kWh = 3.600 × [latex]10^{6}[/latex] J; 1 J = 1 C · V MM for Cr	Information implied:
kilowatt-hours	Asked for:

STRATEGY

1. Convert the mass of Cr to mol [latex]e^{-}[/latex] using stoichiometry

mass Cr [latex]\underrightarrow{MM} mol Cr \underrightarrow{2 mol Cr/12 mol e^{-}} mol e^{-}[/latex]

2. Find kWh by finding the energy in joules (C × V) and then convert to kWh (3.600 × [latex]10^{6}[/latex] J = 1 kWh)

[latex]mol e^{-} \underrightarrow{1 mol e^{-} = 9.648 × 10^{4} C} coulomb (C) \underrightarrow{V}[/latex] C · V → J → kWh

Accepted Answer

ⓐ

6.00 A = 6.00 C/s

6.00[latex]\frac{C}{s}[/latex] × (48 × 60)s × [latex]\frac{1 mol e^{-}}{9.648 × 10^{4}C}[/latex] = 0.179 mol [latex]e^{-}[/latex]

1. mol [latex]e^{-}[/latex]

0.179 mol [latex]e^{-}[/latex] × [latex]\frac{2 mol Cr}{12 mol e^{-}} × \frac{52.00 g Cr}{1 mol Cr}[/latex] = 1.55 g

2. Mass Cr

ⓑ

(0.215 L)(1.25 mol/L)(12 mol [latex]e^{-}[/latex]/mol [latex]Cr_{2}O_{7}^{2-}[/latex]) = 3.225 mol [latex]e^{-}[/latex]	1. mol [latex]e^{-}[/latex]
3.225 mol [latex]e^{-} × \frac{9.648 × 10^{4} C}{1 mol e^{-}}[/latex] = 3.11 × [latex]10^{5}[/latex] C	2. C
time = [latex]\frac{3.11 × 10^{5} C}{6.00 C/s} = 5.16 × 10^{4} s[/latex] = 14.4 h	3. Time

ⓒ

1.00 g Cr × [latex]\frac{1 mol Cr}{52.00 g} × \frac{12 mol e^{-}}{2 mol Cr}[/latex] = 0.115	1. mol [latex]e^{-}[/latex]
0.115 mol [latex]e^{-} × \frac{9.468 × 10^{4} C}{1 mol e^{-}} × 4.5 V × \frac{1 J}{1 C · V} × \frac{1 k Wh}{3.600 × 10^{6} J}[/latex] = 0.014 kWh	2. kWh

END POINTS

1. Note that whether you are given data to determine the moles of a species in a reaction or the coulombs of electricity, you can get to moles of electrons.
2. The value given for the voltage used is irrelevant for parts (a) and (b). You only need it to find the number of kilowatt hours.

(0.215 L)(1.25 mol/L)(12 mol $e^{-}$ /mol $Cr_{2}O_{7}^{2-}$ ) = 3.225 mol $e^{-}$	1. mol $e^{-}$
3.225 mol $e^{-} × \frac{9.648 × 10^{4} C}{1 mol e^{-}}$ = 3.11 × $10^{5}$ C	2. C
time = $\frac{3.11 × 10^{5} C}{6.00 C/s} = 5.16 × 10^{4} s$ = 14.4 h	3. Time

1.00 g Cr × $\frac{1 mol Cr}{52.00 g} × \frac{12 mol e^{-}}{2 mol Cr}$ = 0.115	1. mol $e^{-}$
0.115 mol $e^{-} × \frac{9.468 × 10^{4} C}{1 mol e^{-}} × 4.5 V × \frac{1 J}{1 C · V} × \frac{1 k Wh}{3.600 × 10^{6} J}$ = 0.014 kWh	2. kWh

Question 17.8: Chromium metal can be electroplated from an aqueous solution...

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