Question

Can you please explain, how a reaction is most likely to occur between (assume standard conditions apply) the following?

The link to electrochemical series http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chemdata-w.pdf

page 4

Answer 1

You need to set up the relevant electrochemical half-reactions from the provided Table.

Explanation:

Then, combine them (the metal oxidation and the ionic solution reduction) and see if the answer is positive or negative. A negative value indicates NO reaction is spontaneous. A positive value indicates that the reaction will proceed.

From the Table:
`Ag^+ + e^- ⇌ Ag^0` 0.80V (anode, oxidation)
`Fe^(2+) + 2e^- ⇌ Fe^0` -0.44V (cathode, reduction)

They both need to be set up in the real direction from the "Standard Table" direction (reduction) for the desired reaction:

`Ag^0 ⇌ Ag^+ + e^(-)` -0.80V (this is taken into account in the following formula)

`Fe^(2+) + 2e^- ⇌ Fe^0` -0.44V
`2Ag^0 + Fe^(2+) ⇌ 2Ag^(+) + Fe^0` (2 electrons exchanged)

For the "Standard Table" values the cell is calculated with the following formula:

`E_(cell)^@ = E(cathode)^@ - E_(anode)^@ = -0.44V – 2(0.80)V`
`= -2.04V`

This reaction will NOT proceed.

The same is true for the lead metal.

`Pb^(2+) + 2e^– ⇌ Pb^0` -0.13V
`Pb^0 + Fe^(2+) ⇌ Ag^+ + Fe^0` (2 electrons exchanged)
`-0.44 - (-0.13) = -0.31V`

For C. the chlorine will be oxidized, and the Mn will be reduced.
For D. the peroxide oxygen will be reduced and the iodine will be oxidized.