Question

A voltaic 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.

Answer 1

Answer:

[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°-$\frac{R*T}{n*F}$*ln(Q)

where:

• ΔEº = potential in standard conditions.
• R = gas constant.
• T = absolute temperature (in degrees Kelvin).
• n = number of moles that have participation in the reaction.
• F = Faraday constant (with a value of 96500 C / mol, approx.)
• Q = reaction ratio. For the reaction aA + bB → cC + dD, Q adopts the expression:   Q = $\frac{[C]^{c}*[D]^{d} }{[A]^{a}*[B]^{b} }$ In this case [C] and [D], they refer to the partial pressures, also known as molar concentrations in the case of gases or ions in solution, for the reaction products. [A] and [B] are also partial pressures but in the case of reagents. The exponents refer to the amount of moles that make up each substance that is participating in the reaction. Substances that are in a solid state have a unit concentration (that is, their value is 1), so they do not appear in Q.

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° - $\frac{0.059}{n}$*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:

• ΔE=1.04 V
• ΔE°=0.99 V
• n= 2 electrons
• Q=$\frac{[Sn^{2+}] }{[Hg_{2} ^{+2} ]}$ where [Hg₂²⁺] is 0.24 M

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

1.04 V= 0.99 V - $\frac{0.059}{2}$*ln$\frac{[Sn]^{2+} }{0.24}$

[Sn²⁺]=0.044 M=0.044 mol/L