Question

For the following reaction at room temperature, what is [Zn2+] when Ecell=1.14V?

Zn(s)+Cu²⁺(aq,1.50M)⟶Zn²⁺(aq, ?M)+Cu(s)
anode (oxidation):Zn(s)⟶Zn²⁺(aq)+2e⁻E∘Zn²⁺/Zn=−0.76
Vcathode (reduction):Cu²⁺(aq)+2e−⟶Cu(s)E∘Cu²⁺/Cu=0.34 V

Report your answer with two significant figures.

Answer 1

To determine the concentration of Zn2+ at Ecell=1.14V, the Nernst equation is used incorporating both the standard electrode potentials and the reaction quotient. The standard electrode potential is calculated and then applied within the Nernst equation to solve for [Zn2+].

Step-by-step explanation:

To find the concentration of Zn2+ when Ecell=1.14V for the reaction Zn(s)+Cu2+(aq,1.50M)⟶Zn2+(aq, ?M)+Cu(s), we'll use the Nernst equation which integrates the standard electrode potentials and the reaction quotient (Q).

The standard electrode potential (E°) for the cell is calculated by subtracting the anode potential from the cathode potential:

E° = E°cathode - E°anode
E° = 0.34 V - (-0.76 V)
E° = 1.10 V

Then, we use the Nernst equation to find [Zn2+]:

Ecell = E° - (0.0591/n) × log(Q)
1.14 V = 1.10 V - (0.0591/2) × log([Zn2+]/1.50)

By rearranging and solving for [Zn2+], we can find that answer.