Electrochemistry

Question

Cell Diagram

1. Calculate the e.m.f of the following cells under standard conditions and show cell diagram

a. Ni2+ /Ni and Sn4+/ Sn2+

b. I - /I2 and Ag+ / Ag

2.An electrochemical cell is constructed by connecting an aluminium half-cell

a. Write the cell diagram for the aluminium-silver electrochemical cell

b. Write the half equations and redox equation involve in the reaction

3. Predict either reaction between aqueous bromine with iodide solution are spontaneous reaction or non-spontaneous reaction:

Answer 1

We can't easily draw on this platform, so showing the cell diagram isn't really possible. I'll do my best to describe it.

1a. Ni²⁺ /Ni and Sn⁴⁺/ Sn²⁺

Ni²⁺ + 2e- ==> Ni(s) Eº = -0.25 V (anode, oxidation)

Sn⁴⁺ + 2e- ==> Sn²⁺ Eº = 0.15 (cathode, reduction)

Eº_cell = 0.15 - (-0.25) = 0.40 V

To draw the cell diagram, place the cathode on the right and anode on the left. On the right you have Sn⁴⁺ -> Sn²⁺ and on the left you have Ni(s) -> Ni²⁺. Separate the compartments by a salt bridge.

1b. I - /I₂ and Ag⁺ / Ag

I₂ + 2e- ==> 2I^- Eº = 0.53 V (anode, oxidation)

Ag⁺ + e- ==> Ag(s) Eº = 0.80 V (cathode, reduction)

Eº_cell = 0.80 - 0.53 = 0.27 V

Draw diagram as above but on the right you have Ag⁺ + e- -> Ag(s) and on the left you have 2I^- -> I2

You try the others.