Avoltaic cell consists of a hg/hg22+ electrode (e°= 0.85 v) and a sn/sn2+ electrode (e°= –0.14 v). calculate [sn2+] if [hg22+] = 0.24 m and ecell = 1.04 v at 25°c.

[Sn²⁺]=0.044 M=0.044 mol/L

Explanation:

Standard potential of a cell, E°, is the potential originated when all species are present in standard thermodynamic conditions, that is, 1 M concentrations for solutes in solution and 1 atm for gases.

When conditions are different from the standard, the Nernst equation is used:

ΔE=ΔE°-*ln(Q)

where:

The most used form of this expression, at 25 ° C (298.15 °K), after replacing the numerical value of the constants R and F is:

ΔE= ΔE° - *ln(Q) Equation (A)

The oxidation semi-reaction is one in which there is a loss of electrons, this being the one that has the potential of reduction E ° negative or less.   In this case the couple Sn/Sn²⁺ oxidizes.

The reduction half-reaction is one where there is an electron gain, its reduction potential being positive or greater.  In this case the pair Hg/Hg₂²⁺ is reduced.

Then:

Sn ⇒ Sn²⁺ + 2e⁻  E°=-0.14 V

Hg₂²⁺ + 2e⁻ ⇒ 2 Hg  E°=0.85 V

The net reaction is Sn(s) + Hg₂²⁺ ⇒ Sn²⁺ + 2 Hg(s)

Being ΔE°=E°red - E°ox

Then ΔE°=0.85 V - (-0.14 V)

ΔE°= 0.99 V

Then, at 25 ° C, you know that:

By replacing this in Equation (A) it is possible to determine the value of the concentration [Sn²⁺]:

1.04 V= 0.99 V - *ln

[Sn²⁺]=0.044 M=0.044 mol/L

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