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).