Question

Selective Oxidation The standard reduction potential for the half-reaction Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following cathodic half reactions would produce, at the anode, a spontaneous oxidation of Sn to Sn2+ but not Sn2+ to Sn4+. 2H+ + 2e - H2 Fe3+ + 3e + Fe Sn2+ + 2e Fe2+ + 2e → Pb2+ + 2e + Ni2+ + 2e + Ni Submit Answer Incorrect. Tries 4/8 Previous Tries

Answer 1

The standard reduction potential of the half reactions are -

Sn + + 2e → Sn2+ E° = 0.15V

Sn2+ + 2e + Sn E° = -0.14V

We know that when two half reactions are coupled, the half reaction with the higher reduction potential will undergo reduction, and that with the lower reduction potential will undergo oxidation.

Reduction will occur at cathode (right hand electrode) and oxidation will occur at anode (left hand electrode).

Ecell = Ecathode – Eanode

Condition for reaction - Reaction will only occur when
"sathode > EO
, i.e.
cell > 0

[According to modern conventions, we are considering standard reduction potential only. Oxidation potentials are not considered nowadays.]

Now according to the question, the required half reactions are cathodic, that means it will occur at cathode. So its reduction potential must be greater than that at the anode.

According to the question, there will be spontaneous oxidation of Sn to Sn²⁺.So the condition of redox reaction is fulfilled. So the cathodic half reaction will have E^o > -0.14 V.

Also, it is mentioned that there will be spontaneous oxidation of Sn²⁺ to Sn⁴⁺. So the condition of redox reaction is not fulfilled. So the cathodic half reaction will have E^o < 0.15 V.

So we have to find reactions whose standard reduction potentials are less than 0.15 V but more than -0.14 V.

-0.14 V < E^o < 0.15 V. Only then E^o_cell will be positive and the reaction will occur.

Given:

2H+ + 2e + H EⓇ = 0.00 V

Fe3+ + 3e + Fe Eº = -0.036 V

Sn2+ + 2e + Sn E = -0.14 V

Fe2+ + 2e + Fe E° = -0.44 V

Pb2+ + 2e + Pb Eº = -0.13 V

Ni2+ + 2e + Ni Eº = -0.23 V

The half cell reactions that are satisfying the criteria -0.14 V < E^o < 0.15 V are -

1)

2H+ + 2e + H EⓇ = 0.00 V

2)

Fe3+ + 3e + Fe Eº = -0.036 V

3)

Pb2+ + 2e + Pb Eº = -0.13 V

These 3 are the required answer.

(The reaction will not be considered as it is not more than -0.14 V).

Sn2+ + 2e + Sn E = -0.14 V