Question

1. Given electrode potentials :
Fe3+ + e - Fe2+ ; E° = 0.771 V
1. + 2e - →21; E° = 0.536 V.
Eºcell for the cell reaction
2Fe3+ + 21 → 2Fe2+ + I, is -
(1) (2 x0.771 - 0.536) = 1.006 V
(2) (0.771 - 0.5 x 0.536) = 0.503 V
(3) 0.771 - 0.536 = 0.235 V
(4) 0.536 -0.771 = -0.235 V​

Answer 1

Answer:

can write down the three half reactions:

Fe2++2e−Fe3++e−Fe3++3e−⟶Fe⟶Fe2+⟶FeE1E2E3=xV=yV=zV

The third half-reaction is simply the sum of the first two. However, we cannot just sum E1+E2 in order to obtain E3. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential E in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

x joules per coulomb, or

96500x joules per mole of electrons, or

2×96500x joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

y joules per coulomb,

96500y joules per mole of electrons,

the same 96500y joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

2×96500x+96500y joules per mole of reaction, which is

3×96500z joules per mole of reaction, or

96500z joules per mole of electrons, or

z joules per coulomb.

Now you can simplify this to 2x+y=3z and calculate the potential you want.