A voltaic cell is constructed using Al and Al3+ in one half- cell and Ag with Ag+ in the other half-cell.

Question

A).Write the half-cell reaction that will occur at each electrode.
B).Write the net cell reaction and workout the cell potential
C). Which metals is the anode and which is the cathode.

Answer 1

A) Al → Al3+ + 3e- (oxidation half-cell reaction at anode)
Ag+ + e- → Ag (reduction half-cell reaction at cathode)

B) Al + Ag+ → Al3+ + Ag (net cell reaction)
The standard reduction potentials of Ag+ to Ag and Al3+ to Al are +0.80 V and -1.68 V, respectively. The overall cell potential would be the difference between these two potentials, which is:

Eº cell = Eº cathode - Eº anode
Eº cell = +0.80 V - (-1.68 V)
Eº cell = +2.48 V

C) Al is the anode (where oxidation occurs), and Ag is the cathode (where reduction occurs).

Answer 2

Calculate the standard Gibbs energy of the given reaction at 25 °C
CO(g)+0.5 O2 =CO2

Answer 3

The standard Gibbs energy change (ΔG°) of a reaction can be calculated using the standard Gibbs energy of formation (ΔG°f) of the reactants and products involved in the reaction, as well as the stoichiometric coefficients of the reaction.

The standard Gibbs energy of formation (ΔG°f) is the change in Gibbs energy that occurs when one mole of a substance is formed from its constituent elements in their standard states, which are defined as the most stable form of the element at 1 bar pressure and a specified temperature (usually 25 °C).

Using the values of ΔG°f at 25 °C (298 K) for CO(g) and CO2(g) from standard tables as -137.3 kJ/mol and -394.4 kJ/mol, respectively, we can calculate the standard Gibbs energy change for the given reaction as follows:

ΔG° = ΣnΔG°f (products) - ΣnΔG°f (reactants)

ΔG° = (-394.4 kJ/mol) - [(-137.3 kJ/mol) + (0.5 × 0 kJ/mol)] (where 0 kJ/mol is the ΔG°f of O2(g))

ΔG° = (-394.4 kJ/mol) - (-68.65 kJ/mol)

ΔG° = -325.75 kJ/mol

Therefore, the standard Gibbs energy change (ΔG°) for the given reaction at 25 °C is -325.75 kJ/mol.