Chemistry HW help

Question

You decide to try to make a new kind of battery. After doing some research, you decide your two cells will be the NiO2 / Ni2+ and the H2MoO4/ MoO2 half-cells. This leads to the following electrochemical reaction:

2H+(aq) + NiO2(s) + MoO2(s) —> <—Ni2+ (aq) + H2MoO4 (aq)

E°cell= 0.943V

To test the prototype of your battery, you set up 100.0 mL of each half cell at standard conditions. You allow the battery to run for 5.00 hours at 2.2 A.

1.How many moles of electrons have been used in this reaction during this time?

2. Calculate the new concentrations of the NiO2 / Ni2+ and

H2MoO4 / MoO2 half cells after those 5.00 hours.

3.What is the new voltage given off by the battery after those 5.00 hours?

4.If you let the battery complete run its course (i.e., if the battery dies), how much energy would you need to completely recharge the battery?

5. If you increased the initial concentration of Ni2+, how would that change the INITIAL cell voltage reading before using the battery?

Answer 1

This is a rather lengthy problem, so I'll attempt to get started and maybe others will weigh in also, as I'm a little uncertain about this particular galvanic cell. However, from the initial balanced equation and the positive value for Eº_cell

2H⁺(aq) + NiO₂(s) + MoO₂(s) <==> Ni²⁺ (aq) + H₂MoO₄ (aq)

it looks like NiO₂ is the anode (Ni will be oxidized) and that MoO₂ is the cathode (H⁺ will be reduced). We could write the cell as NiO₂ | Ni²⁺(aq), 1 M || 2H⁺, 1M | MoO₂

(1). The balanced equation shows the reaction uses 2 moles of electrons under standard conditions (1 M).

(2). 5.00 hours x 3600 sec / hr = 18,000 sec

2.2 A = 2.2 C / sec

2.2 C / sec x 18,00 sec x 1 mol e- / 96485 C = 0.410 mols e-

mols H⁺ reduced = 0.410 mol e- x 2 mol H⁺ / 2 mol e- = 0.410 mols H⁺ reduced

mols Ni oxidized = 0.410 mol e- x 1 mol Ni / 2 mol e- = 0.205 mols Ni oxidized

New [H⁺] = 1 M - 0.410 M = 0.590 M

New [Ni²⁺] = 1 M + 0.205 M = 1.205 M

(3). Nernst equation: E_cell = Eº_cell - 0.0592/n x ln Q

E_cell = 0.943 - 0.0296 ln 1.205/0.590

E_cell = 0.922 V

(4). Calculate energy needed to replace all work done using nFEº_cell. This will be a negative value since the cell is doing work on the surroundings.

(5). Use the Nernst equation as in (3) to see the effect of increasing [Ni²⁺]