Question

one half cell of a volytaic cell has a nickel electrode ina 1mol/l nickel (ii) chloride solution. the other half cell has a cadmium electrode in a 1 mol/l cadmium chloride solution. a) find hte cell potential

Answer 1

The cell potential for the voltaic cell with a nickel electrode in a 1 mol/L nickel (II) chloride solution and a cadmium electrode in a 1 mol/L cadmium chloride solution is +0.15 V, indicating a spontaneous reaction that can generate electrical energy.

Step-by-step explanation:

To find the cell potential for the voltaic cell with a nickel electrode in a 1 mol/L nickel (II) chloride solution and a cadmium electrode in a 1 mol/L cadmium chloride solution, we need to look at the standard reduction potentials for both half-reactions.

The standard electrode potentials (E°) for the relevant half-reactions would be:

• Ni²+ + 2e⁻ → Ni(s) with E° = -0.25 V (Nickel)
• Cd²+ + 2e⁻ → Cd(s) with E° = -0.40 V (Cadmium)

The cell potential, E°cell, is calculated by subtracting the lower potential (the anode reaction) from the higher potential (the cathode reaction), thus E°cell = E°cathode - E°anode.

We designate cadmium as the anode because it has a more negative standard reduction potential. Therefore, the cell potential would be:

E°cell = E°(Ni/Ni²+) - E°(Cd/Cd²+) = (-0.25 V) - (-0.40 V) = +0.15 V

Since the cell potential is positive, the reaction is spontaneous, indicating that electrical energy can be produced by this cell.