I NEED HELP PLEASE, THANKS! Electrochemistry is important in many aspects of daily life.
i. Define voltaic cell.
ii. Fill in the blanks for the drawing of a voltaic cell that’s made with copper/copper (II) nitrate (E° = 0.34 V) and zinc/zinc (II) nitrate (E° = –0.76 V). Briefly explain the role of the salt bridge.
iii. Using the equation E°cell = E°cathode – E°anode, calculate the overall cell potential for the cell.
i.
a. _
b. _
c. _
d. _
e. _
f. _
g. _
h. _

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

(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.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o