Question

Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(b) Hg₂²⁺(aq) → Hg²⁺(aq) + Hg(l)

Answer 1

The given reaction is not spontaneous.

We must recognize changes in oxidation states that take place across elements in order to balance these equations. To accomplish this, keep in mind following guidelines:

A neutral element on its own has an oxidation number of zero.For a neutral molecule, the total number of oxidations must be zero.The net charge of an ion is equal to the sum of its oxidation numbers.In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.In a compound with no oxygen present the other halogens will also prefer -1.

One of the mercury atoms is oxidized from +1 to +2 in the simple aqueous ion, for a loss of 1 electron.

Oxidation half-reaction:

$0.5 Hg^{2+} _{2} (aq)$ →$Hg^{2+} (aq) + 1e^-$

$E^o _{ox} = - 0.92 V$

The other mercury is reduced from +1  to zero in mercury metal, for a gain of 1 electron.

Reduction half-reaction:

$0.5 Hg^{2+} _{2} (aq) + 1 e^-$ →$Hg(l)$

$E^o _{red} = 0.85V$

This is a disproportionation redox reaction !

Net reaction:

$Hg^{2+} _{2} (aq)$ →$Hg^{2+} (aq) + Hg (l)$

$E^o _{cell} = 0.85 - 0.92 = -0.07V$

The cell potential is negative so this reaction is NOT spontaneous.

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