Choose the balanced equation for the reaction. Au3+(aq) + Cu+(aq) → Au(s) + Cu2+(aq) Au(s) + Cu2+(aq) → Au3+(aq) + Cu+(aq) Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq) Au(s) + 3Cu2+(aq) → Au3+(aq) + 3Cu+(aq) b Use the standard reduction potentials for the galvanic cell to determine E°.

Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)

1.35 V

Explanation:

Given that the reduction potential of

Cu^2+(aq) + e- > Cu^+(aq) is +0.15 V

While the reduction potential of

Au3+(aq) + 3 e- > Au(s) is + 1.50V

It is clear that the Cu^+(aq)/Cu^2+(aq) system is the anode while Au^3+(aq)/Au(s) system is the cathode based on the reduction potentials shown above. The number of electrons transferred (n) =3

E°cell = E°cathode - E°anode

E°cell= 1.50-0.15

E°cell= 1.35 V

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :