QUESTION: Calculate the standard free-energy change (deltaG°) for the following reaction at 25 °C in kJ.
2Ag^3+(aq) + 3Zn(s)<==> 2Au(s) + 3Zn^2+(aq)
here's what i have so far:
first i wrote out the half reactions:
Au^3+(aq) + 3e^- --> Au(s) = +1.52*2
Zn(s) --> Zn^2+(aq) + 2e^- = +0.76*3
so E°cell= (1.52 * 2) + (0.76 * 3)=5.32V
then i plugged it into the equation..
deltaG°= -nFE°cell, where F=96,485 C/mol
am i on the right track? how do i find n so i can solve for deltaG°?
There is one error, you have to have each equation have the same number of electrons. You did, however, get the E values right.
The correct equations should be like this:
2Au^3+(aq) + 6e^- --> 2Au(s) = +1.52*2
3Zn(s) --> 3Zn^2+(aq) + 6e^- = +0.76*3
"n" refers to the amount of electrons in the reaction. Since now they are equal in both reactions, n = 6.
There is a second error. The Ecell is not 5.32 v.
Ecell = 1.52 + 0.76 = ?
You do NOT multiply by the coefficients 2 and 3. Yes, n will be 6.
Yes, you are on the right track. To find the value of n in the equation ΔG° = -nFE°cell, you need to determine the number of electrons transferred in the overall balanced equation. In this case, you can examine the balanced equation:
2Ag^3+(aq) + 3Zn(s) ⇌ 2Au(s) + 3Zn^2+(aq)
The coefficient of electrons represented in the half-reaction for Au^3+ is 3, and the coefficient in the half-reaction for Zn is 2. Since the two half-reactions are added together, the number of electrons transferred is determined by the least common multiple of these coefficients, which is 6. Therefore, n = 6.
Now, you can substitute the values into the equation:
ΔG° = -nFE°cell
ΔG° = -(6)(96,485 C/mol)(5.32 V)
Calculating this expression will give you the value of ΔG° in Joules. Remember to convert the final answer to kJ by dividing by 1000.