Consider the following cell at standard conditions:
Zn(s) |Zn²⁺(aq) M Fe²⁺(aq) |Fe(s)
Calculate the value of $\Delta$ G_r^$\omicron$for the reaction that occurs when current is drawn from this cell.

Question

Consider the following cell at standard conditions: Zn(s)|Zn²⁺(aq)M Fe²⁺(aq)|Fe(s) Calculate the value of $\Delta$G_r^$\omicron$for the reaction that occurs when current is drawn from this cell.

Quizplus · Accepted Answer

We can start by writing the balanced equation of the cell reaction: Zn(s) + Fe²⁺(aq) -> Zn²⁺(aq) + Fe(s) From the given cell notation, we can see that the anode (oxidation) is Zn and the cathode (reduction) is Fe. The standard reduction potentials for the half-reactions are: Fe3+(aq) + e- -> Fe2+(aq) E° = +0.771 V Zn2+(aq) + 2e- -> Zn(s) E° = -0.762 V The overall cell potential can be calculated using the equation: ∆G° = -nFE°cell where n is the number of electrons transferred in the balanced equation, F is the Faraday constant (96,485 C/mol), and E°cell is the standard cell potential. In this case, n = 2 (2 electrons transferred per balanced equation). E°cell = E°cathode - E°anode = +0.771 V - (-0.762 V) = +1.533 V ∆G° = -2 × 96,485 C/mol × (+1.533 V) = -296,039 J/mol = -296 kJ/mol Since the reaction releases energy (negative ∆G°), the sign of ∆G° in the Gibbs free energy equation is positive: ∆G° = ∆H° - T∆S° At standard conditions, ∆G° = ∆H°, so: ∆H° = +296 kJ/mol The answer in the closest option to this value is A, with ∆H° = +62 kJ/mol. However, this value is positive, implying that the reaction is endothermic and requires an input of energy to occur. This contradicts our initial assumption that the cell is discharging (current is being drawn from it). Therefore, we can conclude that none of the answer choices are correct.

Consider the Following Cell at Standard Conditions: Zn(s)|Zn2+(aq)M Fe2+(aq)|Fe(s) Calculate Δ\DeltaΔ

Consider the Following Cell at Standard Conditions:
Zn(s)|Zn²⁺(aq)M Fe²⁺(aq)|Fe(s)
Calculate $\Delta$