A concentration cell is constructed of two hydrogen electrodes. The pressure of H2 is 1.00 atm in each hydrogen electrode. One of the electrodes is immersed in a solution with [H+] = 1.0 M and the other in a solution with 0.62 M NH3. What will be the value of Ecell? (Hint: First, you need to calculate [H+] for the NH3 solution.)
Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4
To calculate the value of Ecell, we need to first find the concentrations of H+ in the NH3 solution.
NH3 is a weak base and reacts with water to form the ammonium ion (NH4+) and hydroxide ion (OH-):
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium expression for the reaction is:
Kw = [NH4+][OH-] / [NH3]
At 25°C, the value of Kw (ion product constant for water) is 1.0 x 10^-14 M^2.
The concentration of NH3 in the solution is given as 0.62 M. Let's assume x is the concentration of [NH4+] formed and [OH-] would also be x (as they are in a 1:1 ratio).
Therefore, we can substitute these values into the equilibrium expression:
1.0 x 10^-14 = (x)(x) / 0.62
x^2 = 1.0 x 10^-14 * 0.62
x^2 = 6.2 x 10^-15
x = √(6.2 x 10^-15)
x ≈ 7.87 x 10^-8
Now we have the concentration of [NH4+], which is approximately 7.87 x 10^-8 M. Since [NH4+] and [H+] are in a 1:1 ratio (due to the reaction NH3 + H2O ⇌ NH4+ + OH-), the concentration of [H+] in the NH3 solution is also 7.87 x 10^-8 M.
Next, we can calculate the cell potential (Ecell) using the Nernst equation:
Ecell = E°cell - (0.0592 V / n) * log(Q)
In this case, since the hydrogen electrodes are the same, E°cell is assumed to be zero. n represents the number of electrons transferred in the balanced redox equation (which is 2 for the hydrogen electrode).
Q is the reaction quotient, which is given by the ratio of product concentrations to reactant concentrations:
Q = ([H+] in solution with NH3) / ([H+] in solution with H+)
Q = (7.87 x 10^-8 M) / (1.0 M)
Now we can substitute the values into the Nernst equation:
Ecell = 0 - (0.0592 V / 2) * log(7.87 x 10^-8 M / 1.0 M)
Ecell = - (0.0592 V / 2) * log(7.87 x 10^-8)
Calculating this value, we find:
Ecell ≈ -0.0592 V * (-7.10383)
Ecell ≈ 0.4198 V
Therefore, the value of Ecell is approximately 0.420 V (with 3 significant digits).