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Question

A voltaic cell consists of Cu²⁺/Cu and Cd²⁺/Cd half-cells. Each compartment has a volume of 0.500 L with the following initial concentrations: [Cu²⁺] = 1.250 M; [Cd²⁺] = 0.250 M at 25.0 °C. What is the [Cu²⁺] when the voltage has dropped to 0.719 V? (E⁰ for Cu²⁺/Cu = 0.340 V and E⁰ for Cd²⁺/Cd = -0.400 V.)

Answer 1

From the standard reduction potentials, we see that Cu will be the cathode (reductionO, and Cd will be the anode (oxidation). Thus, we can write the following:

Cu²⁺ + 2e- ==> Cu(s) ... Eº = 0.340 V

Cd(s) ==> Cd²⁺ + 2e- ... Eº = -0.400 V

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Cu²⁺ + Cd(s) ==> Cd²⁺ + Cu(s) ... Eº = 0.740 V

For non-standard condition such as those in the current problem, we need to use the Nernst equation to find the Ecell:

Ecell = Eºcell - RT/nF lnQ and @ 25ºC ...

Ecell = Eºcell - 0.0592/n log Q

Ecell = 0.740 - (0.0592/2) log [Cd²⁺]/[Cu²⁺]

Ecell = 0.740 - 0.0296 log 0.250/1.250 = 0.740 + 0.0207

Ecell = 0.761 V

During the operation of the cell, the [Cu²⁺] will decrease and [Cd²⁺] will increase and thus the voltage will change. We are told the new voltage is 0.729 V and we are asked to find [Cu²⁺]. We can view this in a table as follows:

Cu²⁺(aq) + Cd(s) ==> Cu(s) + Cd²⁺(aq)

1.25.............0................0..........0.25..........Initial

-x..............................................+x..............Change

1.25-x........................................0.25+x.......Final

So, Q = [Cd²⁺] / [Cu²⁺] which is 0.25+x / 1.25-x. Inserting this into the equation we have...

0.729 = 0.761 - 0.0296 log [Cd²⁺]/[Cu²⁺]

-0.032 = - 0.0296 log [Cd²⁺]/[Cu²⁺]

log [Cd²⁺]/[Cu²⁺] = 1.081

[Cd²⁺]/[Cu²⁺] = 12.05

0.25+x / 1.25-x = 12.05

x = 1.135

[Cu²⁺] = 1.250 - 1.135

[Cu²⁺] = 0.115 M