Calculate the Gibbs free energy change for the reaction below when initial concentrations of CR3+ and Cu2+ are 1 M (F= 96500 J/V mole e-)
2Cr3+ (aq) + 3Cu (s) --> 2Cr (s) + 3Cu2+ (aq)
I assume we use the deltaG= -nFE equation, but how do you know n when Cr is using 3 e- and Cu is using 2 e-. and how do you find E? which is the cathode/anode?
The value for n depends upon how you solve the problem. If you use the total balanced equation as you have it written, then n is the total number of electrons (6 in this case) exchanged. If you decide to solve it by calculating the half cell potential for EACH, then adding the two half cells to obtain the total reaction (in which case you DON'T multiply Cu E by 3 and Cr E by 2) then n is 2e for the copper half and 3e for the Cr half. As for the anode/cathode part,
if you work using the entire equation, if E is +, then Cr will be the anode (anode is where oxidation occurs and Cr is losing electrons which makes it oxidized) but if E is -, the Cu will be the anode and Cr the cathode.
623
Ah, the wonderful world of electrochemistry! Let's break it down, shall we?
To determine the value of n in the equation ΔG = -nFE, we need to consider the stoichiometric coefficients of the electrons involved during the redox reaction. In this case, Cr3+ is using 3 electrons, while Cu2+ is using 2 electrons. So, n can be either 3 or 2, depending on which species we choose to focus on.
Now, to find E, we use the standard reduction potentials (E°) of each species. The reduction potential of a single species describes its tendency to accept or donate electrons relative to the standard hydrogen electrode (SHE). By comparing the reduction potentials of the two half-reactions involved, we can calculate the overall potential of the cell.
As for determining which is the cathode (positive electrode) and anode (negative electrode), we look at the reduction potentials. The species with the more positive reduction potential will be reduced (cathode), while the species with the less positive (or even negative) reduction potential will be oxidized (anode).
So, let's find the relevant reduction potentials and determine the values of n, E, and the nature of the cathode and anode!
To calculate the Gibbs free energy change (ΔG) for the given reaction, you are correct in using the equation ΔG = -nFE, where ΔG is the Gibbs free energy change, n is the number of moles of electrons transferred, F is Faraday's constant (96,500 J/V∙mol e-), and E is the cell potential or electromotive force of the reaction.
To determine the value of n, you need to examine the balanced chemical equation. In this case, the balanced equation shows that 2 moles of Cr3+(aq) are reduced to form 2 moles of Cr(s) by the transfer of 3 moles of electrons per mole of Cr3+. Similarly, 3 moles of Cu2+(aq) are reduced to form 3 moles of Cu(s) by the transfer of 2 moles of electrons per mole of Cu2+. Since the number of electrons transferred is different for each species, you can consider them separately.
For Cr3+:
- The reduction of 1 mole of Cr3+ requires the transfer of 3 moles of electrons.
- Thus, n for the reduction of Cr3+ is 3.
For Cu2+:
- The reduction of 1 mole of Cu2+ requires the transfer of 2 moles of electrons.
- Thus, n for the reduction of Cu2+ is 2.
Now, to calculate the overall value of n, you need to consider the stoichiometry of the balanced equation. In this case, since 2 moles of Cr3+ are reduced and 3 moles of Cu2+ are reduced, the larger value of n should be used. Therefore, n for this reaction is 3.
Next, to find E, you need to calculate the cell potential or electromotive force of the reaction. This can be determined using standard electrode potentials (E°) for each half-reaction and the Nernst equation, or by using experimental data if available.
The standard reduction potential, E°, of each half-reaction can be found in tables. For the reduction of Cr3+(aq) to Cr(s), the standard reduction potential (E°) can be determined, and similarly, for the reduction of Cu2+(aq) to Cu(s), the E° value can be found.
To decide which half-reaction occurs at the anode (oxidation) and which occurs at the cathode (reduction), you need to compare the reduction potentials (E°) for the half-reactions. The half-reaction with the more positive (or less negative) reduction potential will occur at the cathode, while the one with the less positive (or more negative) reduction potential will occur at the anode.
Once you determine the anode and cathode reactions, you can calculate the overall cell potential (Ecell) by subtracting the anode potential from the cathode potential. This Ecell value can then be used in the ΔG = -nFE equation to calculate the Gibbs free energy change (ΔG) for the given reaction.
It's important to note that the values used for E° should be for the given temperature and under standard conditions (1 M concentration, 1 bar pressure, and 298 K). If the conditions are not at standard state, you may need to apply the Nernst equation to adjust the E° values.
I hope this explanation helps you understand how to determine the value of n and how to find E for the given reaction. Feel free to ask any further questions!