Calculate [Pb2+] in a voltaic cell

Question

A voltaic cell consists of an Mn/Mn²⁺ half-cell and a Pb/Pb²⁺ half-cell. Calculate [Pb²⁺] when [Mn²⁺] is 2.7 M and E_cell = 0.49 V.

Answer 1

This question examines a voltaic (galvanic) cell), which means that the overall chemical reaction is spontaneous, and the cell potential is positive. From a table of standard reduction potentials, we can see that Pb²⁺ is easier to reduce than Mn²⁺ is (each ion is being reduced to its respective metal), because the lead ion reduction has a more positive reduction potential than the manganese ion reduction:

Pb²⁺ + 2e- --> Pb(s) E°red = -0.13 V

Mn²⁺ + 2e- --> Mn(s) E°red = -1.18 V

In a galvanic cell, this then means that Pb²⁺ + 2e- --> Pb(s) will operate in the forward direction (reduction; cathode 1/2 reaction), and that Mn²⁺ + 2e- --> Mn(s) will operate in reverse (oxidation of Mn(s); anode 1/2 reaction), for the overall reaction:

Pb²⁺ + 2e- --> Pb(s) E°red (cathode) = -0.13V

Mn(s) --> Mn²⁺ + 2e- E°ox (anode) = +1.18 V

(Note we change the sign and designation of E° for the anode reaction when we reverse it to an oxidation)

OVERALL REACTION:

Pb²⁺ + Mn(s) --> Mn²⁺ + Pb(s) with n = 2 (the number of electrons exchanged in the overall reaction)

The standard cell potential is given by E°cell = E°red + E°ox = 1.05 V (as expected this is positive for a galvanic cell)

The concentration of Mn²⁺ ions is 2.7 M in this experiment (not standard), and the actual cell potential measured is 0.49 V. We can use the Nernst equation to determine cell potentials under non-standard conditions, and that is what we use here given the non-standard potential to solve for the unknown concentration of Pb²⁺. The Nernst equation requires a temperature, which is not specified here, so we will assume to be 25 °C or 298 K.

Nernst equation:

Ecell = E°cell - RT/nF ln Q, or

Ecell = E°cell - 0.0592/n log Q at 25 °C (combining the numerical values for the ideal gas constant and Faraday's constant, as well as T=298K, and converting the natural log to the base₁₀ log)

For our balanced equation, n = 2 and Q = [Mn²⁺]/[Pb²⁺]

Rearranging the equation to isolate Q, we get the following:

Ecell - E°cell = -0.0592/2 V log Q;

log Q = -(Ecell - E°cell)*2/0.0592

log Q = -(0.49 V - 1.05 V) * (2/0.0592 V)

log Q = 18.91

Q is then = 10^18.91 -- WOW! this is a HUGE number!!! Notice that relatively small voltage changes correspond to very large changes in the magnitude of the reaction quotient. This is due BOTH to the log term and the small pre-multiplier term (0.0592/n V).

We are finally ready to solve for [Pb²⁺]...

Q = 10^18.91 or 8.297 x 10¹⁸, which also = [Mn²⁺]/[Pb²⁺]

Solving for [Pb²⁺] = [Mn²⁺]/8.297 x 10¹⁸

[Pb²⁺] = 2.7 M/8.297 x 10¹⁸

[Pb²⁺] = 3.3 x 10^-19 M

It's helpful to do 2 things to check this answer:

1. plug the concentrations of both ions back into the Nernst equation and see if you obtain the actual voltage given from the E°cell and Q. CHECK, Ecell = 0.49 V
2. reason whether your answer makes sense... Ecell here is LESS positive than E°cell, so it is less spontaneous under these conditions than under standard conditions, when Q = 1. By Le Châtelier's principle, a reaction will be less spontaneous in the forward direction when Q > 1 and more spontaneous when Q < 1. Actual conditions here make the reaction less spontaneous (Ecell < E°cell), and this is consistent with the huge value of Q obtained in our calculation.

Answer 2

From a table of standard reduction potentials:

Mn²⁺ + 2e- ==> Mn(s) ... Eº = -1.18 V (anode)

Pb²⁺ + 2e- ==> Pb(s) ... Eº = -0.13 V (cathode)

Overal reaction:

Pb²⁺(aq) + Mn(s) ==> Pb(s) + Mn²⁺(aq)

Eº_cell = -0.13 + 1.18 = 1.05 V

To find [Pb²⁺] when the Ecell is 0.49 V and [Mn²⁺] is 2.7 M, we must uses the Nernst equation. However, this requires us to know the temperature at which the cell is operating, and that information is not provided. I will assume it is at 25ºC.

Nernst Equation:

Ecell = Eºcell - RT/nF ln Q and at 25ºC, this simplifies to

Ecell = Eºcell - 0.0592/n log Q where Q = [Mn²⁺] / [Pb²⁺]

0.49 = 1.05 - 0.0592/2 log Q

0.56 = 0.0296 log Q

log Q = 18.9

Q = 8.3x10¹⁸ = [Mn²⁺] / [Pb²⁺]

8.3x10¹⁸ = 2.7 / [Pb²⁺]

[Pb²⁺] =3.3x10¹⁹ M

(be sure to check the math)