Given that E^o = 0.52 V for the reduction of Cu+(aq) + e - (yields) Cu(s)
Calculate E^o, (delta)G^o and K for the following reaction at 25°C.
2Cu2+(aq) (yields) Cu2+(aq) + Cu(s)
That's actually word for word from the problem. That's why I've been having so much difficulty with it. The answers were given as 0.37 V, -36 kJ, and k = 2x10^6. I just want to understand how it's possible.
I think you have made a typo. It makes the difference between a half reaction and a complete reaction. Note that the charges don't balance. I see +4 on the left and +2 on the right.
I THINK what you meant to type is
2Cu^+(aq) ==> Cu^2+ + Cu(s)
Why did the copper ion go to therapy? Because it had too many Cu2+ thoughts!
To calculate the standard cell potential (E^o), we can use the Nernst equation. The reduction half-reaction is Cu+(aq) + e- -> Cu(s), with E^o = 0.52 V. So, the standard reduction potential for Cu2+(aq) + 2e- -> Cu(s) would be E^o = 2 * 0.52 V = 1.04 V.
To calculate the standard free energy change (ΔG^o), we can use the equation ΔG^o = -nF E^o. Here, n is the number of moles of electrons transferred (2 for Cu2+ + 2e- -> Cu), and F is the Faraday constant (9.6485 x 10^4 C/mol). Thus, ΔG^o = -2 * 9.6485 x 10^4 C/mol * 1.04 V = -2 * 9.6485 x 10^4 J/mol.
The equilibrium constant K can be calculated using the equation ΔG^o = -RT ln K, where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (25 + 273 = 298 K). Plugging in the values, we get -2 * 9.6485 x 10^4 J/mol = -8.314 J/(mol·K) * 298 K * ln K. Solving for K, we find K = e^(-2 * 9.6485 x 10^4 J/mol / (8.314 J/(mol·K) * 298 K)).
To calculate E^o, delta G^o, and K for the given reaction, we will use the Nernst equation and the relationship between delta G^o and K.
1. Calculate E^o for the given reaction:
E^o = E^o(red) - E^o(ox)
Given: E^o(red) = 0.52 V (reduction potential for Cu+ to Cu)
E^o(ox) = 0 V (since Cu is the standard electrode for the reaction)
E^o = 0.52 V - 0 V = 0.52 V
Therefore, E^o for the given reaction is 0.52 V.
2. Calculate delta G^o for the given reaction:
delta G^o = -nF E^o
Given: n = number of moles of electrons transferred
In the given reaction, 2 moles of electrons are transferred. (Cu2+(aq) gains 2 electrons to form Cu(s))
F = Faraday's constant = 96485 C/mol (at 25°C)
delta G^o = -2 * 96485 C/mol * 0.52 V
delta G^o = -100336.8 J/mol
Therefore, delta G^o for the given reaction is -100336.8 J/mol.
3. Calculate K for the given reaction using the relationship between delta G^o and K:
delta G^o = -RT ln(K)
Given: R = gas constant = 8.314 J/(mol K)
T = temperature = 25°C = 298 K
delta G^o = -100336.8 J/mol
R = 8.314 J/(mol K)
T = 298 K
-100336.8 J/mol = -8.314 J/(mol K) * 298 K * ln(K)
Solving for ln(K):
ln(K) = -100336.8 J/mol / (-8.314 J/(mol K) * 298 K)
ln(K) = 44.58
Taking the exponential of both sides to get rid of the natural logarithm:
K = e^(44.58)
Therefore, K for the given reaction is approximately 3.82 x 10^19.
To calculate E^o, (delta)G^o, and K for the given reaction, we will use the following equations:
1. Nernst equation: E^o = E^o(reduction) - E^o(oxidation)
2. Gibbs free energy equation: (delta)G^o = -nFE^o
3. Equilibrium constant equation: (delta)G^o = -RT ln(K)
Step 1: Calculate E^o(oxidation)
E^o(oxidation) is the reverse of the reduction half-reaction, so we flip the reduction half-reaction equation:
Cu(s) (yields) Cu+(aq) + e-
E^o(oxidation) = -E^o(reduction) = -0.52 V
Step 2: Calculate E^o
Using the Nernst equation, we can calculate E^o:
E^o = E^o(reduction) - E^o(oxidation)
E^o = 0.52 V - (-0.52 V)
E^o = 1.04 V
Step 3: Calculate (delta)G^o
To calculate (delta)G^o, we need to know the number of electrons transferred (n) in the reaction. In this case, the number of electrons transferred is 1.
(delta)G^o = -nFE^o
(delta)G^o = -(1)(Faraday's constant)(E^o)
(delta)G^o = -(1)(96500 C/mol)(1.04 V)
(delta)G^o = -100060 J/mol
Step 4: Calculate K
To calculate K, we use the equation:
(delta)G^o = -RT ln(K)
First, convert the temperature to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, substitute the values:
-100060 J/mol = -(8.314 J/mol·K)(298.15 K) ln(K)
Solve for ln(K):
ln(K) = (-100060 J/mol) / [(8.314 J/mol·K)(298.15 K)]
ln(K) = -1.261
Take the exponential of both sides:
e^(ln(K)) = e^(-1.261)
K = 0.283
Therefore, the value of E^o is 1.04 V, the value of (delta)G^o is -100060 J/mol, and the value of K is 0.283 for the given reaction at 25°C.