Multiple Choice
What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10, according to the Nernst Equation?
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20. Electrochemistry
Cell Potential: The Nernst Equation
Multiple Choice
If [Br–] = 0.010 M and [Al3+] = 0.022 M, predict whether the following reaction would proceed spontaneously as written at 25ºC:
Al (s) + Br2 (l) ⇌ Al3+ (aq) + Br– (aq)
Standard Reduction Potentials
Al3+ (aq) + 3 e– → Al (s) E°red = –1.66 V
Br2 (l) + 2 e– → Br– (aq) E°red = +1.09 V
A
2.90 V
B
2.60 V
C
3.03 V
D
2.75 V
Verified step by step guidance1
Identify the half-reactions involved in the given redox reaction. The reduction half-reaction for Al3+ is: Al3+ (aq) + 3 e– → Al (s) with E°red = –1.66 V. The reduction half-reaction for Br2 is: Br2 (l) + 2 e– → Br– (aq) with E°red = +1.09 V.
Determine the oxidation and reduction processes. Since Al is being oxidized from Al (s) to Al3+ (aq), reverse the Al3+ reduction half-reaction to get the oxidation half-reaction: Al (s) → Al3+ (aq) + 3 e–. The E° for this oxidation reaction is the negative of the reduction potential: E°ox = +1.66 V.
Calculate the standard cell potential (E°cell) by adding the standard reduction potential of the reduction half-reaction to the standard oxidation potential of the oxidation half-reaction: E°cell = E°red (Br2) + E°ox (Al).
Use the Nernst equation to determine the cell potential under non-standard conditions: Ecell = E°cell - (RT/nF) * ln(Q), where Q is the reaction quotient. For this reaction, Q = [Al3+]^1 / [Br–]^2.
Substitute the given concentrations into the expression for Q: Q = (0.022) / (0.010)^2. Calculate the value of Q and then use it in the Nernst equation to find the cell potential at 25ºC. Compare this potential to zero to determine if the reaction is spontaneous (Ecell > 0) or non-spontaneous (Ecell < 0).
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