Question

1. An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e−
Red: MnO−4(aq, 1.50 M )+4H+(aq, 1.8 M )+3e−→
MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.

2.A concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.16 V at 25 ∘C.

What is the ratio of the Sn2+ concentrations in the two half-cells?

Express your answer using two significant figures.

3.Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the item to a power source and submerging it into a Au3+ solution. The item itself serves as the cathode, at which the Au3+ions are reduced to Au(s). A piece of solid gold is used as the anode and is also connected to the power source, thus completing the circuit.

What mass of gold is produced when 8.80 A of current are passed through a gold solution for 18.0 min ?

Express your answer with the appropriate units.

Answer 1

1. Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e−     E⁰anode = -0.126
Red: MnO−4(aq, 1.50 M )+4H+(aq, 1.8 M )+3e−→ MnO2(s)+2H2O(l)   E⁰cathode = 0.59

E0cell = E⁰cathode - E⁰anode = 0.59 - (-0.126) = 0.716 Volt

Ecell = E⁰cell - 0.0592 / n log Q

The balanced equation will be:

3Pb + 2MnO4- + 8H+ --> 3Pb+2 + 2MnO2(s) + 4H2O

So n = 6   ; Q = [Pb⁺²]³ / [MnO4-]² [H+]⁸ = (0.18)³ / (1.5)² (1.8)⁸ = 2.352 X 10^-5

Ecell = 0.716 - 0.0098 log (2.352 X 10^-5 ) = 0.716 - (-0.045) = 0.761 V

2.

Sn_(s) --> Sn²⁺_(aq) + 2e^- ; 0.14V oxidation half reaction

Sn²⁺_(aq) + 2e^- --> Sn_(s); -0.14V reduction half cell reaction

Sn_(s) + Sn²⁺_(aq) Þ Sn²⁺_(aq) + Sn_(s)   (overall reaction)

Nernst equation will be

E_cell = E°_cell - 0.0592 /n ● log Q

n= 2

0.16 = 0 - 0.0296 log (Sn+2 / Sn+2)

-5.405 = log (Sn+2 / Sn+2)

[Sn+2] / [Sn+2] = 3.93 X 10^-6

3. What mass of gold is produced when 8.80 A of current are passed through a gold solution for 18.0 min ?

For each mole fo Au+3 the amount of charge = 3 Faraday

Let us calculate the charge applied

Charge = Current X time (seconds) = 8.8 X 18 X 60 = 9504 coloumbs

96500 coloumbs = 1 Faraday

so 9504 coloumbs = 9504 / 96500 Faraday = 0.0984 moles

so moles of Au+3 reduced will be 0.0984 / 3 = 0.0328 moles

Atomic mass of gold = 197 g / moles

so mass of 0.0328 moles = 0.0328 X 197 = 6.461 grams