The value of Delta G degree_rxn for the reaction Au^3+(aq) + 3Cu(s) rightarrow A
ID: 519296 • Letter: T
Question
The value of Delta G degree_rxn for the reaction Au^3+(aq) + 3Cu(s) rightarrow Au(s) + 3Cu^+(aq) is -283 kJ/mol. What is the value of E degree _cells in volts, for this reaction at 25.0 degree C? Enter your answer to the hundredths place. At 25.0 degree C, the galvanic cell constructed between the half reactions below has K 8.63 times 10^13. What is the value of E degree_cell, in volts, for this reaction? Enter your answer to the thousandths place. HClO_2(aq) + 3H^+(aq) + 4e^- Equilibrium Cl^-(aq)+ 2H_2O(l) H_2O_2(aq) + 2H^+ (aq) + 2e^- Equilibrium 2H_2O(l) The two half reactions below represent the reduction of water under different pH conditions, though the two reactions will give the same E_cell regardless of the pH. What is the E_cell, in volts, for either half-reaction at 25.0 degree c when the pH = 9.15 and the P_H_2(g) = 1.00 atm? Enter your answer to the thousandths place. Remember to include the sign! (For extra fun, peform the calculation using both equations!) 2H^+(aq) + 2e^- rightarrow H_2(g) E degree_red = 0.0000V 2H_2O(l) + 2e^- rightarrow H_2(g) + 2OH^-(aq) E^+_red = -0.8277 V Vanadium metal can be produced by electrolysis of V_2O_5. How many grams of vanadium metal can be produced using a current of 75.0 A over the course of 3.00 hours? Enter your answer to the tenths place.Explanation / Answer
Q10
K = 8.63*10^13
n = 4 electrons are being transferred
so
dG = -RT*ln(K)
dG = -nFE°cell
nFE°cell = RT*ln(K)
E° = RT/(nF)*ln(K)
E° = 8.314*298/(4*96500)*ln(8.63*10^13)
E° = 0.20596 V
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