Question

What potential, in volts, is developed by the following electrochemical cell at 298 K? Al(s) | Al3+(0.00100 M) || Cu2+(0.00100 M) | Cu(s) A13+(aq) + 3e– = Al(s) E°red = –1.66V Cu2+(aq) + 2e– = Cu(s) E°red = +0.34 V (Hint: balance the cell reaction before you proceed.) Question 5 options: 2.01 1.95 2.00 1.97 2.07

Answer 1

The potential developed by the electrochemical cell is 2.00 volts.

Step-by-step explanation:

To find the potential developed by the electrochemical cell, we need to balance the cell reaction first. The balanced cell reaction is:

2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

Next, we can calculate the cell potential, Eocell, using the Nernst equation:

Eocell = Eored(cathode) - Eored(anode)

Substituting the given standard reduction potentials, we get:

Eocell = 0.34 V - (-1.66 V) = 2.00 V

Therefore, the potential developed by the electrochemical cell is 2.00 volts.

Answer 2

The potential developed by the electrochemical cell is approximately 1.97 volts.

Step-by-step explanation:

The given electrochemical cell consists of an Al(s) electrode in a solution of Al³+(0.00100 M) on the left side of the cell notation and a Cu²+(0.00100 M) solution with a Cu(s) electrode on the right side. To find the potential of the cell, we need to balance the reduction half-reactions and calculate the cell potential using the Nernst Equation.

First, we balance the half-reactions:

Al³+(aq) + 3e⁻ → Al(s) (multiply by 2)

Cu²+(aq) + 2e⁻ → Cu(s)

Next, we can use the Nernst Equation: Ecell = E°cell - (0.0592/3)log(Q), where Q is the reaction quotient.

Using the given concentrations, we can calculate Q. The reaction quotient for this cell is Q = [Al³+]²/[Cu²+]. Substituting the values into the Nernst Equation, we can find the cell potential, which is approximately 1.97 volts.