Question

Consider the following reaction proceeding at 298.15 K:

2Al(s)+3Cu²⁺(aq,1.70 M)⟶2Al³⁺(aq, 0.00560 M)+3Cu(s)
If the standard reduction potential for copper(II) is 0.340 V and the standard reduction potential for aluminum is −1.66 V, what is the cell potential for this cell, in volts?

Answer 1

The cell potential for the reaction between aluminum and copper(II) ions is 2.00 V, which indicates that the reaction is spontaneous under standard conditions. This is calculated using the standard reduction potentials of copper(II) and aluminum.

Step-by-step explanation:

To calculate the cell potential for the reaction 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), you need to use the standard reduction potentials for copper(II) and aluminum. The standard reduction potential for copper(II), Cu²⁺/Cu, is given as 0.340 V and for aluminum, Al³⁺/Al, it's given as −1.66 V. Remember, the cell potential (ΔV) is given by the difference between the reduction potentials of the cathode and anode.

Since copper is being reduced, it serves as the cathode with a potential of 0.340 V. Aluminum is being oxidized, serving as the anode, and its potential must be reversed when plugging into the equation (since we are provided with the reduction potential, but oxidation is occurring), giving us +1.66 V.

So the cell potential will be:

ΔV = E°(cathode) - E°(anode) = 0.340 V - (-1.66 V) = 2.00 V.

The positive cell potential indicates that the reaction is spontaneous under standard conditions, according to the electrochemical series.