What is the e°cell for the cell represented by the combination of the following half-reactions? clo4–(aq) + 8h+(aq) + 8e– cl–(aq) + 4h2o(l) e° = 1.389 v , vo2+(aq) + 2h+(aq) + e– vo+(aq) + h2o(l) e° = 0.991 v

The E°cell for the cell represented by the combination of the given half-reactions is 0.398 V

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

This is the type of reaction that occurs in this case.

ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

VO₂⁺(aq) + 2 H⁺(aq) + e⁻ ⇔ VO⁺(aq) + H₂O(l) E° = 0.991 V

In this case both are written as reductions, and their E ° values as well. The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the first half-reaction expressed.  Therefore, to obtain a reaction, the second semi-reaction must be reversed to be an oxidation, maintaining its constant value. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = 0.991 V

E°cell=Ereduction - Eoxidation

E°cell=1.389 V - 0.991 V

E°cell=0.398 V

Then the E°cell for the cell represented by the combination of the given half-reactions is 0.398 V.

Another way of thinking is that, by inverting the second semi-reaction to be an oxidation, the value of E ° is reversed in the sign, unlike the previous case in which it was constant. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = -0.991 V

In this case:

E°cell=Ereduction + Eoxidation=

E°cell=1.389 V + (-0.991 V)=1.389 V-0.991 V

E°cell=0.398 V

Note that the result obtained is the same. This indicates that either of the two ways proposed is correct, and you will use the one that is most comfortable for you.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.