Answer:
Step-by-step explanation:
To determine the concentration of Cu2+ ions in the given sample, we can use the Nernst equation and the given standard reduction potentials.
First, let's consider the given cell:
Cathode half-cell: Cu2+(aq) + 2e- → Cu(s) (reduction)
Anode half-cell: Fe(s) → Fe2+(aq) + 2e- (oxidation)
The standard cell potential (E°cell) is given as 0.78V. The standard reduction potential for the Cu2+/Cu half-reaction (E° Cu2+/Cu) is not provided directly, but we can calculate it using the E°cell value and the standard reduction potential for the Fe2+/Fe half-reaction (E° Fe2+/Fe).
E°cell = E° Cu2+/Cu - E° Fe2+/Fe
0.78V = E° Cu2+/Cu - (-0.44V)
0.78V = E° Cu2+/Cu + 0.44V
E° Cu2+/Cu = 0.78V - 0.44V
E° Cu2+/Cu = 0.34V
Now, let's use the Nernst equation:
Ecell = E°cell - (0.0592V/n) * log(Q)
Where:
Ecell = cell potential (0.78V)
E°cell = standard cell potential (0.78V)
n = number of electrons transferred in the balanced equation (2 in this case)
Q = reaction quotient
For the given cell, Q can be calculated using the concentrations of Cu2+ and Fe2+:
Q = [Cu2+]/[Fe2+]
Since the concentration of Fe2+ is given as 0.5M, we need to determine the concentration of Cu2+.
Substituting the given values into the Nernst equation:
0.78V = 0.34V - (0.0592V/2) * log([Cu2+]/0.5)
0.44V = (0.0592V/2) * log([Cu2+]/0.5)
Now, we can solve for [Cu2+]:
log([Cu2+]/0.5) = (0.44V * 2) / 0.0592V
log([Cu2+]/0.5) = 14.86
Using the logarithmic property of exponents:
[Cu2+]/0.5 = 10^14.86
[Cu2+] = 0.5 * 10^14.86
Therefore, the concentration of Cu2+ ions in the given sample is approximately 5.011 * 10^14 M.
Please note that this calculation assumes that the activity coefficients and other factors that can affect the cell potential have been neglected. It is always a good practice to consider these factors for accurate results in real-life scenarios.