A voltaic cell is constructed using Al and Al3+ in one half- cell and Ag with Ag+ in the other half-cell.
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.
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).
Calculate the standard Gibbs energy of the given reaction at 25 °C
CO(g)+0.5 O2 =CO2
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.
To answer your questions, let's break down each part step by step:
A) Writing the half-cell reactions:
1. Al and Al3+ half-cell: The anode (oxidation) reaction occurs at the aluminum electrode, where aluminum atoms lose electrons to form aluminum ions:
Al(s) -> Al3+(aq) + 3e-
2. Ag and Ag+ half-cell: The cathode (reduction) reaction occurs at the silver electrode, where silver ions gain electrons to form silver atoms:
Ag+(aq) + 1e- -> Ag(s)
B) Writing the net cell reaction and calculating cell potential:
The net cell reaction is obtained by combining the two half-cell reactions. Since the Al electrode loses electrons (oxidation) and the Ag electrode gains electrons (reduction), we need to multiply the Ag half-cell reaction by 3 to balance the electrons. Then we can sum up the two half-cell reactions:
3Ag+(aq) + 3e- -> 3Ag(s) (multiplied by 3)
Al(s) -> Al3+(aq) + 3e-
Summing both reactions gives us the net cell reaction:
3Ag+(aq) + Al(s) -> 3Ag(s) + Al3+(aq)
To calculate the cell potential (E°cell), you need to know the standard reduction potentials for the half-cell reactions. The given information should include the standard reduction potentials (E°red) for Ag+ to Ag and Al3+ to Al. By looking up these values, you can subtract the smaller E°red from the larger E°red.
E°cell = E°red(cathode) - E°red(anode)
C) Determining the anode and cathode:
In a voltaic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Therefore, based on the half-cell reactions we wrote in part A, the anode is the electrode where the Al(s) is being oxidized to Al3+(aq), and the cathode is the electrode where Ag+(aq) is being reduced to Ag(s).