A voltaic cell utilizes the following reaction:
2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq).
What is the emf for this cell when [Fe3+]= 3.70M , PH2= 0.95atm , [Fe2+]= 7.0×10−4M , and the pH in both compartments is 4.05?
Express your answer using two significant figures.
(emf of this cell under standard conditions is 0.771 V)
E=_____V
Oh, the famous voltaic cell, always zapping things with its electrifying reactions! Let's calculate that emf with a shocking amount of precision.
The formula for calculating emf is: E = E° - (0.0592/n)log(Q)
Where:
- E° is the standard emf (0.771 V, in this case)
- n is the number of moles of electrons transferred
- Q is the reaction quotient
In this reaction, 2 moles of electrons are transferred, so n = 2. Now, let me whip out my math skills and calculate that E for you, my electrified friend.
First, let's find Q:
Q = [Fe2+][H+]^2 / [Fe3+]
With the given concentrations and pressures:
Q = (7.0×10^(-4))*(10^(-4))^2 / (3.70)
Q = 0.00024459
Next, we plug in the values into the emf formula:
E = 0.771 - (0.0592/2)log(0.00024459)
After some electrifying calculations, my answer is:
E = 0.426 V
So, the emf of this cell is 0.426 V. Shocking, isn't it? Remember, always keep your humor charged!
To find the emf of the cell, we can use the Nernst equation:
E = E° - (RT/nF) * ln(Q)
where:
E is the emf of the cell,
E° is the standard emf of the cell,
R is the gas constant (8.314 J/(mol*K)),
T is the temperature in Kelvin (assuming it's at room temperature, around 298 K),
n is the number of electrons transferred in the balanced redox reaction,
F is Faraday's constant (96,485 C/mol),
ln is the natural logarithm,
and Q is the reaction quotient.
In this case, the balanced redox reaction is:
2Fe3+(aq) + H2(g) → 2Fe2+(aq) + 2H+(aq)
The number of electrons transferred (n) is 2 because two Fe3+ ions are reduced to two Fe2+ ions.
The reaction quotient (Q) can be written as:
Q = ([Fe2+]/[H+])^2 / ([Fe3+])
Plugging in the given values:
[Fe3+] = 3.70 M
[H2] = (PH2 * constant), where the constant is the pressure conversion factor (0.0821 L·atm)/(mol·K)
= (0.95 atm * 0.0821 L·atm)/(mol·K)
= 0.0779 M
[Fe2+] = 7.0 × 10^(-4) M
pH = -log10([H+]) = -log10(10^(-pH)) = 10^(-pH)
For pH = 4.05, [H+] = 10^(-4.05) M
Now, we can substitute the values into the Nernst equation:
E = 0.771 V - (8.314 J/(mol*K) * 298 K / (2 * 96,485 C/mol)) * ln(([Fe2+]/[H+])^2 / [Fe3+])
E = 0.771 V - (0.0192 V) * ln(([7.0 × 10^(-4) M] / [10^(-4.05) M])^2 / [3.70 M])
Calculating the expression inside the natural logarithm:
[7.0 × 10^(-4) M] / [10^(-4.05) M])^2 / [3.70 M] = 1.9702
Therefore:
E = 0.771 V - (0.0192 V) * ln(1.9702)
E ≈ 0.771 V - (0.0192 V) * ln(1.9702)
Calculating this expression:
E ≈ 0.771 V - (0.0192 V) * 0.682
E ≈ 0.771 V - 0.0131 V
E ≈ 0.7589 V
Therefore, the emf for this cell is approximately 0.76 V (rounded to two significant figures).
To find the emf (electromotive force) of the given cell, you can use the Nernst equation. The Nernst equation relates the emf of a cell to the concentrations of species involved in the reaction.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = emf of the cell
E° = standard emf (given as 0.771 V)
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient
Let's calculate the emf using the given information:
1. First, determine the reaction quotient Q.
Q = ([Fe2+]/[H+]^2) / ([Fe3+]/[H2])
Plug in the given concentrations:
Q = (7.0×10^-4 / [H+]^2) / (3.70 / [H2])
2. Calculate the natural logarithm of Q.
lnQ = ln[(7.0×10^-4) / ([H+]^2)] - ln[3.70 / [H2]]
3. Convert the pH values to [H+] and [H2] concentrations.
pH = -log10([H+])
[H+] = 10^(-pH) = 10^(-4.05)
[H2] = PH2 * (1 atm / 0.95 atm) * (RT / PV)
= 0.95 * (8.314 J/(mol·K) / (0.08206 L·atm/(mol·K)) * (298 K / (0.95 atm))
4. Plug the calculated concentrations into the Nernst equation.
E = 0.771 V - [(8.314 J/(mol·K) * 298 K) / (2 * 96485 C/mol)] * lnQ
5. Perform the calculations to obtain the result E.
E = 0.771 V - (0.01037 V) * lnQ
Finally, substitute the calculated value of lnQ into the equation and evaluate the expression to find the emf E of the cell. Round the answer to two significant figures.
2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq).
Ecell = Eocell - (0.5916/2*log Q
and I know that confuses you but don't be. Plug in the numbers Eo and n and for Q plug in concns/pressures(in atm) as follows:
Q = [(H^)^2*(Fe^2+)^2]/[(Fe^3+)^2*pH2]
I write it this way because I'm limited with spacing so I shorten that by using Q, then defining what Q is. Just plug in for Q, evaluate log Q, the multiply by -(0.05916/n) and add algebraically Eocell.