Aluminum metal reacts with zinc(ii) ion in an aqueous solution by the following half-cell reactions: al(s) → al3+(aq) + 3e− zn2+(aq) + 2e− → zn(s) a. predict the potential of the cell under standard conditions. (4 points) b. predict whether the reaction will occur spontaneously, or whether a source of electricity will be required for the reaction. justify your answer. (2 points) c. in terms of the metals involved, predict the direction electrons will flow in the reaction. (2 points) d. predict which electrode will lose mass and which will gain mass. (2 points)

We have to predict the potential, whether reaction is spontaneous or not, direction of electron flow, gain/loss of mass at electrode.

The correct answers are:

The reactions are Al(s) → Al³⁺(aq) + 3e⁻  and Zn²⁺(aq) + 2e⁻ → Zn(s). The standard electrode potential for Al/Al³⁺ is +1.18 volts and that of Zn²⁺/Zn is -0.76 volts.Zn²⁺/Zn cathode and Al/Al³⁺  is anode.

(a) Cell potential= Cathode potential - Anode potential

Cell potential= E°(Zn²⁺/Zn) - E°(Al/Al³⁺)

Cell potential= (-0.76- 1.18) volts= -1.94 volts.

(b) As E° value is negative, ΔG° is positive. Hence, the reaction will not occur spontaneously.

(c) Electron will flow from anode to cathode.

(d) Anode (Al/Al³⁺) will lose mass and cathode (Zn²⁺/Zn) will gain mass.

43.452

step-by-step explanation:

284 times .153 = 43.452