Question

A concentration cell is constructed at 25∘C based on the Zn/Zn2+ electrode. The concentration of Zn2+ is known to be 1.8M at the positive electrode. If the cell voltage is initially 0.1016 V, what is the concentration at the negative electrode? Data: Zn2++2e−→ZnE∘=−0.76 V Round your answer to two significant figures.

Answer 1

The concentration of Zn^2+ at the negative electrode in a concentration cell is found by using the Nernst equation, rearranging it to solve for Q, and using the given cell voltage along with the standard reduction potential for Zn/Zn^2+.

Step-by-step explanation:

To calculate the concentration of Zn2+ at the negative electrode of a concentration cell with a known cell voltage, we use the Nernst equation. The standard reduction potential (E°) for the Zn/Zn2+ couple is -0.76 V, and we are given that the cell voltage (Ecell) is initially 0.1016 V. At 25°C, the Nernst equation can be written as:

Ecell = E° + (0.0592/n) * log(Q)

Where Ecell is the cell voltage, E° is the standard electrode potential, n is the number of moles of electrons per mole of reactions (which is 2 for the Zn2+/Zn couple), and Q is the reaction quotient, which is the ratio of the concentrations of the products over the reactants. In this case, Q = [Zn2+] at the negative electrode / 1.8 M at the positive electrode.

By rearranging the Nernst equation and solving for Q, we can calculate the unknown concentration of Zn2+ at the negative electrode.

The concentration of Zn2+ at the negative electrode is calculated as follows:

0.1016 V = -0.76 V + (0.0592/2) * log(Q)

log(Q) = (0.1016 V + 0.76 V) * 2 / 0.0592

Q = 10(0.1016 V + 0.76 V) * 2 / 0.0592

After calculating Q, we find the concentration of Zn2+ at the negative electrode by multiplying Q by 1.8 M.

Answer 2

The concentration of the negative electrode is 1.8 *
`10^{-29` M.

What is the Nernst equation?

The Nernst equation is particularly important in the field of electrochemistry and plays a key role in understanding the behavior of ions in solutions and across cell membranes.

We can know that from the Nernst equation;

E =Eo - 0.0592/n logQ

Eo = −0.76 V

0.1016 = −0.76 - 0.0592/2 log [Red]/[1.8]

0.1016 + 0.76 = - 0.0592/2 log [Red]/[1.8]

0.86 = -0.0296 log [Red]/[1.8]

log [Red]/[1.8] = 0.86/ -0.0296

log [Red]/[1.8] = -29.1

[Red]/[1.8] = Antilog(-29.1)

[Red]/[1.8] = 1 *
`10^{-29`

[Red] = 1 *
`10^{-29` * [1.8]

= 1.8 *
`10^{-29` M