If the standard reduction potential for the half-reaction Fe³⁺ + 3e⁻ → Fe is +0.77 V, what is the potential when [Fe³⁺] = 0.001 M?
Practice Questions
1 question
Q1
If the standard reduction potential for the half-reaction Fe³⁺ + 3e⁻ → Fe is +0.77 V, what is the potential when [Fe³⁺] = 0.001 M?
0.77 V
0.70 V
0.60 V
0.50 V
E = E° - (0.059/3)log(1000) = 0.77 - 0.059/3 * 3 = 0.70 V.
Questions & Step-by-step Solutions
1 item
Q
Q: If the standard reduction potential for the half-reaction Fe³⁺ + 3e⁻ → Fe is +0.77 V, what is the potential when [Fe³⁺] = 0.001 M?
Solution: E = E° - (0.059/3)log(1000) = 0.77 - 0.059/3 * 3 = 0.70 V.
Steps: 8
Step 1: Identify the standard reduction potential (E°) for the half-reaction, which is given as +0.77 V.
Step 2: Determine the number of electrons transferred in the half-reaction. Here, 3 electrons (3e⁻) are involved.
Step 3: Use the Nernst equation to find the potential (E) at a specific concentration. The Nernst equation is E = E° - (0.059/n) log([oxidized species]/[reduced species]).
Step 4: In this case, [Fe³⁺] = 0.001 M, so we need to find the concentration ratio. The reduced form (Fe) is at standard conditions, which is 1 M. Therefore, the ratio is [Fe³⁺]/[Fe] = 0.001/1 = 0.001.
Step 5: Calculate the logarithm of the concentration ratio: log(0.001) = -3.
Step 6: Substitute the values into the Nernst equation: E = 0.77 - (0.059/3) * (-3).
Step 7: Simplify the equation: E = 0.77 + 0.059 = 0.70 V.
Step 8: The final answer for the potential when [Fe³⁺] = 0.001 M is 0.70 V.
