Multiple Choice
What is the emf of this cell when [Fe²⁺] = 1.5 M, [Fe³⁺] = 1.4 × 10⁻² M, P(O₂) = 0.56 atm, and the pH of the solution in the cathode compartment is 3.5, given that the standard cell potential (E°) is 1.23 V?
735
views
20. Electrochemistry
Cell Potential: The Nernst Equation
Multiple Choice
Determine [Fe2+] for the following galvanic cell at 25ºC if given [Sn2+] = 0.072 M, [Fe3+] = 0.0219 M, and [Sn4+] = 0.00345 M.
Sn2+ (aq) + 2 Fe3+ (aq) ⇌. Sn4+ (aq) + 2 Fe2+ (aq) Ecell = + 0.68 V
Standard Reduction Potentials
Sn4+ (aq) + 2 e– →. Sn2+ (aq) E°red = + 0.151 V
Fe3+ (aq) + e– → Fe2+ (aq) E°red = + 0.771 V
A
0.13150 M
B
0.00968 M
C
0.08180 M
D
0.01936 M
Verified step by step guidance1
Identify the half-reactions and their standard reduction potentials: Sn4+ + 2e– → Sn2+ with E°red = +0.151 V and Fe3+ + e– → Fe2+ with E°red = +0.771 V.
Calculate the standard cell potential (E°cell) using the formula E°cell = E°cathode - E°anode. Here, the cathode is the Fe3+/Fe2+ couple and the anode is the Sn4+/Sn2+ couple.
Use the Nernst equation to relate the cell potential (Ecell) to the concentrations of the species involved: Ecell = E°cell - (RT/nF) * ln(Q), where Q is the reaction quotient.
Determine the reaction quotient (Q) for the cell reaction: Q = ([Sn4+][Fe2+]^2) / ([Sn2+][Fe3+]^2). Substitute the given concentrations into this expression.
Rearrange the Nernst equation to solve for [Fe2+], using the given Ecell value and the calculated Q. This will involve algebraic manipulation to isolate [Fe2+] on one side of the equation.
Related Videos
Related Practice
