Multiple Choice
In an electrochemical cell, if the Gibbs free energy change (ΔG) is positive, what can be said about the cell potential (E°cell)?
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20. Electrochemistry
Cell Potential and Gibbs Free Energy
Multiple Choice
Given the reaction and standard cell potentials: Ni(s) + 2Ag+ (aq) ⇌ Ni2+ (aq) + 2Ag(s), where E°Ag+/Ag = +0.80 V and E°Ni2+/Ni = -0.26 V, calculate the standard free energy change (ΔG°) and the equilibrium constant (K) at 25°C for the reaction.
A
ΔG° = +204 kJ/mol, K = 1.2 x 10-35
B
ΔG° = -204 kJ/mol, K = 1.2 x 1035
C
ΔG° = -204 kJ/mol, K = 1.2 x 10-35
D
ΔG° = +204 kJ/mol, K = 1.2 x 1035
Verified step by step guidance1
First, identify the half-reactions and their standard reduction potentials. The given half-reactions are: Ag+ + e- → Ag with E° = +0.80 V and Ni2+ + 2e- → Ni with E° = -0.26 V.
Calculate the standard cell potential (E°cell) for the overall reaction using the formula: E°cell = E°cathode - E°anode. Here, Ag is the cathode and Ni is the anode.
Use the Nernst equation to relate the standard cell potential to the standard free energy change (ΔG°) with the formula: ΔG° = -nFE°cell, where n is the number of moles of electrons transferred, F is Faraday's constant (96485 C/mol), and E°cell is the standard cell potential.
Calculate the equilibrium constant (K) using the relationship between ΔG° and K: ΔG° = -RT ln(K), where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K for 25°C). Rearrange to solve for K.
Ensure the signs and magnitudes of ΔG° and K are consistent with the spontaneity of the reaction. A negative ΔG° indicates a spontaneous reaction, and a large K value suggests the reaction favors the formation of products.
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