electrochemistry

A VCE chemistry student sets up a galvanic cell using two standard half-cells with half reactions.

Half-cell 1: Cr³⁺(aq)+e^-→Cr²⁺(aq)

Half-cell 2: Cr(s)→Cr²⁺(aq)+2e^-

Suitable materials for the electrodes of the two half-cells are

• Half-cell 1: Platinum; Half-cell 2: Platinum

• Half-cell 1: Platinum; Half-cell 2: Chromium

• Half-cell 1: Chromium; Half-cell 2: Chromium

• Half-cell 1: Chromium; Half-cell 2: Platinum

Which of the species strong oxidizing agent?

Cl₂(g) + 2e → 2Cl^-(aq) E^ocell = + 1.36 V

Cu²⁺(aq) + 2e → Cu(s) E^ocell = + 0.34 V

Ni²⁺(aq) + 2e → Ni(s) E^ocell = – 0.25 V

Ca²⁺(aq) + 2e → Ca(s) E^ocell = – 2.87 V

• Cl₂(g)

• Cu²⁺(aq)

• Ni²⁺(aq)

• Ca²⁺(aq)

Which is correct to write cell notation

Anode: Zn

Cathode: Ni

• Zn²⁺ (aq,1M) / Zn (s) // Ni (s) / Ni²⁺ (aq,1M)

• Zn (s) /Zn²⁺ (aq,1M) // Ni²⁺ (aq,1M) / Ni (s)

• Zn²⁺ (s,1M) / Zn (aq) // Ni (aq) / Ni²⁺ (s,1M)

• Zn (aq) /Zn²⁺ (s,1M) // Ni²⁺ (s,1M) / Ni (aq)

Calculate standard cell potential for this equation:

Zn²⁺(aq) + 2e → Zn(s) Eº = – 0.76 V

Cd²⁺(aq) + 2e → Cd(s) Eº = – 0.40 V

• E^ocell = -0.36 V

• E^ocell = -1.16 V

• E^ocell = +1.16 V

• E^ocell = +0.36 V

Based on cell notation below, give the one electrolyte can be used in cathode?

Zn (s) /Zn²⁺ (aq,1M) // H⁺ (aq,1M) / H₂ (g,1 atm)/graphite (s)

• ZnSO₄(aq)

• H₂(aq)

• ZnCl₂(aq)

• HCl(aq)

Based on the cell notation below, give the suitable X can be used ?

X / H₂(g,1 atm)/ H⁺ (aq,1M) // Ag⁺ (aq,1M) //Ag (s)

• Ag (s)

• H (s)

• Pt (s)

• Zn (s)

Which of the pairs give the highest value of voltage cell?

Al³⁺(aq) + 3e → Al(s) E^ocell = - 1.66 V

Cu²⁺(aq) + 2e → Cu(s) E^ocell = + 0.34 V

Ni²⁺(aq) + 2e → Ni(s) E^ocell = – 0.25 V

Br₂(l) + 2e → 2Br^-(aq) E^ocell = + 1.07 V

• Al and Ni

• Cu and Br₂

• Al and Br₂

• Cu and Ni

Arrange the species increasing order oxidizing agent?

Cl₂(g) + 2e → 2Cl^-(aq) E^ocell = + 1.36 V

Cu²⁺(aq) + 2e → Cu(s) E^ocell = + 0.34 V

Ni²⁺(aq) + 2e → Ni(s) E^ocell = – 0.25 V

Ca²⁺(aq) + 2e → Ca(s) E^ocell = – 2.87 V

• Ca < Ni < Cu < Cl^-

• Cl^- < Cu < Ni < Ca

• Ca²⁺ < Ni²⁺ < Cu²⁺ < Cl₂

• Cl_{2 <} Cu²⁺ < Ni²⁺ < Ca²⁺

How to increase cell potential, Ecell based on this Nernst equation:

• Increase temperature

• Decrease concentration Ni²⁺

• Increase concentration Ni²⁺

• Choose E^ocell anode and cathode the positive value

State one functions of salt bridge

• to maintain the movement of electron

• to allowing electron to flow and complete the circuit

• to separate two electrolyte

• to maintain the electrical neutrality of the cell