Predicting the products of electrolysis & Using E values

In electrolytic cell that uses aqueous copper(II) sulfate as electrolyte and copper as electrodes, what is the product at the anode?

• Oxygen gas and water

• Copper (II) ions

• Copper metal

• Hydrogen gas

What is the gas produced at the cathode during the electrolysis of sodium chloride solution?

• water

• hydrogen

• oxygen

• sodium

What ions are present in molten aluminium oxide (Al2O3)?

• Al³⁺, O^2-, H⁺, OH^-

• Al³⁺, OH^-

• Al³⁺, O^2-

A certain galvanic cell has its spontaneous cell reaction:

Zn + Sn²⁺ → Zn²⁺ + Sn

Which is the half-reaction occurring at the anode?

• Zn → Zn²⁺ + 2e-

• Zn²⁺ + 2e- → Zn

• Sn²⁺ + 2 e- → Sn

• Sn → Sn²⁺ + 2 e-

Given these reduction potential data:

Mg²⁺ + 2e- → Mg = -2.36V

Cu²⁺ + 2e- → Cu = +0.34V

Which one is the spontaneous reaction when the two half-cells are combined?

• Mg2+ + Cu²⁺ → Mg + Cu

• Mg²⁺ + Cu → Mg + Cu²⁺

• Mg + Cu → Mg²⁺ + Cu²⁺

• Mg + Cu²⁺ → Mg²⁺ + Cu

Manganese (Mn) has more negative standard potential compared to nickel (Ni). Which of the following is correct when Mn cell and Ni cell are combined to form galvanic cell (spontaneous reaction)?

• Mn is placed at cathode

• Mn is oxidized at cathode

• Ni is placed at cathode

• Ni is oxidized at anode

For this reaction: Zn + Sn²⁺ → Zn²⁺ + Sn

What happens when the concentration of tin ions is increased?

• The reaction will be non-spontaneous

• The reaction will be more spontaneous

• The cell potential will be more negative

• The reaction will be less spontaneous

For this reaction: Fe + 3Ag⁺ → Fe³⁺ + 3Ag

What happens when the concentration of iron (III) ions is increased?

• The reaction will be non-spontaneous

• The reaction will be more spontaneous

• The cell potential will be more positive

• The reaction will be less spontaneous

What reaction occurs at the anode in the galvanic cell?

Ag⁺/Ag = 0.80V

Ni²⁺/Ni = -0.25V

• Ag⁺ + e^- → Ag

• Ni²⁺ + 2e^- → Ni

• Ag → Ag⁺ + e^-

• Ni → Ni²⁺ + 2e^-

Fe³⁺ + e^- → Fe²⁺

Calculate the electrode potential at 298 K if the concentration of Fe²⁺ is 0.02 mol dm^–3

and the concentration of Fe³⁺ is 0.1 mol dm^–3.

• E = E^o - 0.059/1 log[Fe³⁺]/[Fe²⁺]

• E = E^o - 0.059/2 log[Fe³⁺]/[Fe²⁺]

• E = E^o - 0.059/2 log[Fe²⁺]/[Fe³⁺]

• E = E^o - 0.059/1 log[Fe²⁺]/[Fe³⁺]