Chemistry 30 Electrochemistry

Print out the practice set Word RTF PDF

Print out the answers Word RTF PDF

For questions 1 to 3, two half-cells are connected under standard conditions to make an electrochemical cell. Use the Table of Standard Reduction Potentials included with this assignment to obtain the half-reactions involved. For each:

a. write the equation for each half-reaction that will occur

b. label each half-reaction as oxidation or reduction

c. calculate the voltage of the electrochemical cell

d. the net overall balanced redox equation.

e. diagram the cell, clearly indicating the following

the electrodes in appropriate electrolytic solutions

label each electrode as anode or cathode

label each electrode as positive post or negative post

diagram the flow of electrons through the external circuit

a salt bridge with appropriate electrolytic solution

flow of ions from the salt bridge to the two half-cells

1. iron-iron(II) ion (Fe|Fe²⁺) and lead-lead(II) ion (Pb|Pb²⁺)

			E°(V)
cathode	reduction	Pb²⁺ + 2e^- → Pb_(s)	-0.13	+ electrode
anode	oxidation	Fe_(s) → Fe²⁺ + 2e^-	+0.45	- electrode
balanced net equation		Fe_(s) + Pb²⁺ → Pb_(s) + Fe²⁺	+0.32	total voltage



2. chromium-chromium(III) ion (Cr\|Cr³⁺) and rubidium-rubidium ion (Rb\|Rb⁺)

		balance for e^-		E° (V)
cathode	reduction		Cr³⁺ + 3e^- → Cr_(s)	-0.74	+ electrode
anode	oxidation	× 3	Rb_(s) → Rb⁺ + e^-	+2.98	- electrode
balanced net equation		3 Rb_(s) + Cr³⁺ → Cr_(s) + 3Rb⁺		+2.24	total voltage

3. copper-copper(I) ion (Cu|Cu⁺) and aluminum-aluminum ion (Al|Al³⁺)

(NOTE: Be sure to use the Cu¹⁺ half-reaction, not Cu²⁺)

balance for e^-

E° (V)

cathode

reduction

× 3

Cu⁺ + e^- → Cu_(s)

+0.52

+ electrode

anode

oxidation

Al_(s) → Al³⁺ + 3 e^-

+1.66

- electrode

balanced net
equation

Al_(s) + 3 Cu⁺ → 3Cu_(s) + Al³⁺

+2.18

total
voltage

An electrochemical cell is created using gold and magnesium half-cells.

Determine which half-cell will undergo oxidation and which will undergo reduction,

If the mass of the magnesium electrode changes by 5.0 g, what will be the change in mass of the gold electrode, and will its mass increase or decrease?

Part a.

		balance for e^-		E° (V)
cathode	reduction	× 2	Au³⁺ + 3e^- → Au_(s)	+1.50	+ electrode
anode	oxidation	× 3	Mg_(s) → Mg²⁺ + 2 e^-	+2.37	- electrode
balanced net equation		3Mg_(s) + 2Au³⁺ → 2Au_(s) + 3Mg²⁺		+3.87	total voltage

Part b.

The balanced equation is required to answer part b:

3Mg_(s) + 2Au³⁺ → 2Au_(s) + 3Mg²⁺

From this equation we see that 3 moles of Mg react with 2 moles of Au to give us the ratio:

3 Mg

2 Au

Since the question concerns mass, we will need to know the molar masses of each element, in order to convert between moles and mass. Molar masses are equivalent to atomic masses, found in the periodic table. The unit for molar mass is g/mol. (Need a review? Check here)

molar mass of Mg: 24.3 g ·mol^-1 molar mass of Au: 197.0 g ·mol^-1

The question gives us the mass of Mg and asks us to find mass of Au. Set up the equation so that all units will cancel except g Au (mass of Au):

g Au	=	5.0 g Mg	×	1 mol Mg 24.3 g	×	197.0 g mol Au	×	2 Au 3 Mg	=	27.0 g Au
										answer

The mass of gold will change by 27.0 g.

Our half-reactions tell us that gold ions combine with gold ions to produce solid gold. Thus our gold electrode will increase it’s mass by 27.0 g.

Return to Notes

Chemistry 30

Redox Reactions & Electrochemistry Practice Answers

Teacher Resources

Other Links

Print out the practice set	Word	RTF	PDF
Print out the answers	Word	RTF	PDF