What is the copper(II)-ion concentration at 25°C in the cell Zn(s)| Zn²⁺(aq,1.0 M)|| Cu²⁺(aq)| Cu(s)if the measured cell potential is 1.01 V? The standard cell potential is 1.10 V. A) 0.030 M B) 0.99 M C) 0.00091 M D) 1.0 M E) 0.99M

Question

Quizplus · Accepted Answer

To solve the problem, we can use the Nernst equation:

Ecell = E°cell - (RT/nF) ln(Q)

where E°cell is the standard cell potential, R is the gas constant, T is temperature, n is the number of moles of electrons transferred in the balanced equation, F is Faraday's constant, and Q is the reaction quotient.

We can start by writing the balanced equation for the cell reaction:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

The cell diagram gives us the information about the species involved in the anode and cathode compartments of the cell. We can label the anode half-cell as Zn(s) | Zn2+(aq) and the cathode half-cell as Cu2+(aq) | Cu(s). The vertical lines mean a phase boundary, and the double vertical line means a salt bridge.

We can then write the expression for Q:

Q = [Zn2+(aq)] / [Cu2+(aq)]

We are given the measured cell potential (Ecell = 1.01 V) and the standard cell potential (E°cell = 1.10 V). We can find the difference between them:

ΔE°cell = E°cell,cathode - E°cell,anode

where E°cell,cathode is the standard reduction potential of the cathode half-reaction and E°cell,anode is the standard oxidation potential of the anode half-reaction. We can use the standard reduction potentials from a table:

Cu2+(aq) + 2e- → Cu(s)  E°red = +0.34 V
Zn2+(aq) + 2e- → Zn(s)  E°red = -0.76 V

ΔE°cell = E°red,cathode - (-E°red,anode) = 1.10 V - (-0.76 V) = 1.86 V

The number of moles of electrons transferred in the cell reaction is 2, since two electrons are transferred per ion.

We can now substitute the values into the Nernst equation:

1.01 V = 1.10 V - (0.0257 V/n) ln(Q)
ln(Q) = -0.0346
Q = e^(-0.0346) = 0.966

We can solve for [Cu2+] using the expression for Q:

Q = [Zn2+(aq)] / [Cu2+(aq)]
0.966 = [Zn2+(aq)] / [Cu2+(aq)]

We know that the initial concentration of Cu2+ is 1.0 M. We can solve for [Zn2+]:

[Zn2+(aq)] = 0.966 [Cu2+(aq)] = 0.966 M

Finally, we can solve for [Cu2+] using the equation above:

0.966 = [Zn2+(aq)] / [Cu2+(aq)]
[Cu2+(aq)] = [Zn2+(aq)] / 0.966 = 0.00091 M

Therefore, the answer is C, 0.00091 M.

What Is the Copper(II)-Ion Concentration at 25°C in the Cell