Question

For the following reaction at room temperature, what is [Zn2+] when Ecell ​=1.05 V ?

Zn(s)+Cu²⁺(aq,0.00500M)⟶Zn²⁺(aq, ?M + Cus(s)
anode (oxidation) :Zn(s)⟶Zn²⁺(aq)+2e⁻ EZn2+/Zn′​=−0.760
cathode (reduction) : Cu²⁺(aq)+2e−⟶Cu(s)​?M)+Cu(s) VECu²/Cu​=0.340 V​
Report your answer with three significant figures.

Answer 1

To find the concentration of Zn2+ ions when Ecell is 1.05 V, one must use the Nernst equation and standard electrode potentials to calculate the reaction quotient Q and solve for [Zn2+].

Step-by-step explanation:

The question at hand involves finding the concentration of Zn2+ ions in solution when the cell potential (Ecell) is given as 1.05 V for the reaction of zinc metal with copper ions. To solve this, we must use the Nernst equation, which relates the cell potential to the standard electrode potential and the reaction quotient (Q).

The standard cell potential (Eocell) for the reaction can be calculated via the formula:

Eocell = Eocathode - Eoanode

To find the reaction quotient Q, which is [Zn2+]/[Cu2+], we will use the Nernst equation:

Ecell = Eocell - (RT/nF) * ln(Q)

At room temperature (25°C, which is 298 K), and for the reaction involving the transfer of 2 moles of electrons (n=2), the Nernst equation simplifies to:

Ecell = Eocell - (0.0591 V/n) * log(Q)

Thus, by substituting the known values and solving for [Zn2+], we can find the concentration of zinc ions when the cell potential is 1.05 V.

Steps:

1. Calculate Eocell by subtracting Eoanode from Eocathode.
2. Rearrange the Nernst equation to solve for Q.
3. Calculate [Zn2+] using the value of Q and the given [Cu2+].