Question

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Answer 1

Answer:

Here's what I get  

Explanation:

(i) Voltaic cell

A voltaic cell is a device that uses a chemical reaction to produce electrical energy.

(ii) Overall Cell Potential

The standard reduction potentials for the half-reactions are

                               ℰ°/V

Cu²⁺ + 2e⁻ ⇌ Cu    0.34

Zn²⁺ + 2e⁻ ⇌ Zn   -0.76

The half-reaction with the more positive potential is the reduction half-reaction. It is the reaction that occurs at the cathode.

The half-reaction with the more negative potential is the oxidation half-reaction. It is the  reaction that occurs at the anode.

We reverse that half-reaction and subtract the voltages to get the cell reaction.

                                                          ℰ°/V

Cathode:  Cu²⁺ + 2e⁻ ⇌ Cu              0.34

Anode:     Zn ⇌ Zn²⁺ + 2e⁻              -0.76

Cell:          Zn +  Cu²⁺ ⇌ Zn²⁺ + Cu    1.10

$\mathcal{E}_{\text{cell}}^{\circ} = \mathcal{E}_{\text{cat}}^{\circ} - \mathcal{E}_{\text{an}}^{\circ} = \text{0.34 V} - \text{(-0.76 V)} = \text{0.34 V} + \text{0.76 V} = \textbf{1.10 V}$

(iii) Diagram

The specific labels will depend on your textbook.

They are often as follows.

a. Electron flow

b. Voltmeter or lightbulb

c. Electron flow

d. Cathode or Cu

e. Cu²⁺(aq) and NO₃⁻(aq)

f. Salt bridge

g. Zn²⁺(aq) and NO₃⁻(aq)

h. Anode or Zn

The salt bridge enables ions to flow in the internal circuit and to maintain electrical neutrality in the two compartments.

It often consists of a saturated solution of KCl.

As Zn²⁺ ions form in the anode compartment, Cl⁻ ions move in to provide partners for them.

As Cu²⁺ ions are removed from the cathode compartment, K⁺ ions move in to replace them.