CALCULATE DELTA G FOR REACTION
Cu2(aq)+2Ag(s) gives Cu(s)+2Ag(aq)
Given,
E0 Ag+/Ag=0.80 v and E0 Cu2+/Cu=0.34 V
To calculate the ΔG (change in Gibbs free energy) for a reaction, we can use the equation:
ΔG = -nFE
Where:
ΔG is the change in Gibbs free energy
n is the number of moles of electrons transferred
F is the Faraday constant (96485 C/mol)
E is the cell potential (in volts)
First, we need to determine the number of moles of electrons transferred in the reaction. Looking at the balanced equation:
Cu2(aq) + 2Ag(s) → Cu(s) + 2Ag(aq)
We can see that 2 moles of electrons are transferred per mole of reaction.
Now, we can calculate the ΔG using the given standard reduction potentials:
E° Ag+/Ag = 0.80 V
E° Cu2+/Cu = 0.34 V
To determine the overall cell potential (E° cell), we can use the equation:
E° cell = E° cathode - E° anode
Since Ag(s) is the cathode (reduction half-reaction) and Cu2+(aq) is the anode (oxidation half-reaction), we have:
E° cell = E° Ag+/Ag - E° Cu2+/Cu
E° cell = 0.80 V - 0.34 V
E° cell = 0.46 V
Next, we can substitute the values into the equation for ΔG:
ΔG = -nFE
Since n = 2 (as mentioned earlier) and F = 96485 C/mol, we have:
ΔG = -2 * 96485 C/mol * 0.46 V
Finally, we can calculate the value of ΔG:
ΔG = -88781 J/mol
Therefore, the value of ΔG for the given reaction is -88781 J/mol.
To calculate ΔG (Gibbs free energy) for a reaction, we can use the equation:
ΔG = -nFE
where:
- ΔG is the change in Gibbs free energy,
- n is the number of electrons transferred in the reaction,
- F is the Faraday constant (~96,485 C/mol), and
- E is the standard cell potential (also known as E°).
In this particular reaction, we have the following half-reactions:
1) Ag+(aq) + e- → Ag(s) (reduction half-reaction)
2) Cu2+(aq) + 2e- → Cu(s) (oxidation half-reaction)
Now, let's calculate the overall cell potential (E° cell) for the reaction by taking the difference between the standard reduction potentials of the two half-reactions:
E° cell = E° reduction (Ag+) + E° oxidation (Cu2+)
Given that E° Ag+/Ag = 0.80 V and E° Cu2+/Cu = 0.34 V, we have:
E° cell = 0.80 V + (-0.34 V)
E° cell = 0.80 V - 0.34 V
E° cell = 0.46 V
Since this is a spontaneous reaction, the number of electrons transferred (n) is equal to the stoichiometric coefficients of electrons in the balanced equation. In this case, 2 electrons are transferred, as shown in the balanced equation:
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
Now, we can calculate ΔG using the equation mentioned earlier:
ΔG = -nFE
= -(2 mol)(96,485 C/mol)(0.46 V)
Simplifying this expression will give you the value of ΔG for the reaction.
Eorxn = -0.8 + 0.34 = -0.46
dG = -RT*ln*K