In a certain experiment, with chemical equation Zn(s) + H+
(aq) → Zn2+(aq) +
H2(g), the emf (E) of the cell is found to be 0.45 V at 25°C. Suppose that [Zn2+]
= .8 M and PH2 = 1.0 atm. Calculate the molar concentration of H+
.
To calculate the molar concentration of H+, we can use the Nernst equation. The Nernst equation relates the cell potential (E) to the concentrations of the reactants and products involved in the redox reaction. The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the cell potential
- E° is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced chemical equation
- F is Faraday's constant (96485 C/mol)
- ln is the natural logarithm
- Q is the reaction quotient
First, let's identify the values we have:
- E = 0.45 V
- T = 25°C = 298 K
- E°, R, and F are constants that we can look up in tables
- n = 2 (since 2 electrons are transferred in the balanced chemical equation)
Now, we need to calculate the reaction quotient (Q). The reaction quotient is given by the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.
Q = ([Zn2+]/[H+])^2 * (PH2)^0
Substituting the given values:
Q = (0.8/[H+])^2 * (1.0)^0
Q = (0.8/[H+])^2
Next, we can rearrange the Nernst equation to solve for [H+]:
ln([Zn2+]/[H+]) = [nF/(RT)] * (E° - E)
Substituting the known values:
ln(0.8/[H+]) = [2 * 96485 C/mol / (8.314 J/(mol·K) * 298 K)] * (E° - 0.45 V)
Now, calculate the right side:
ln(0.8/[H+]) = 0.231 * (E° - 0.45 V)
Next, we can rearrange the equation to solve for [H+]:
0.231 * (E° - 0.45 V) = ln(0.8/[H+])
Now, solve for [H+]:
[H+] = 0.8 / exp(0.231 * (E° - 0.45 V))
To calculate the molar concentration of H+, you'll need to use the values of E° for the Zn2+/Zn half-reaction and the temperature-dependent constant R in the calculation.