What is the [Cu²⁺] when the voltage has dropped to 0.729 V? (E⁰ for Cu²⁺/Cu = 0.340 V and E⁰ for Cd²⁺/Cd = -0.400 V.)

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 °C. What is the [Cu²⁺] when the voltage has dropped to 0.729 V? (E⁰ for Cu²⁺/Cu = 0.340 V and E⁰ for Cd²⁺/Cd = -0.400 V.)

Answer 1

First things first: we need to find the cathode (reduction) and the anode (oxidation). We do this by looking at the standard reduction potentials:

Cu²⁺ ==> Cu ... Eº = 0.340 V

Cd²⁺ ==> Cd ... Eº = -0.400

So Cu will be the cathode and Cd will be the anode

Cu²⁺(aq) + Cd(s) ==> Cu(s) + Cd²⁺(aq) ... Eºcell = 0.340 - (-0.400) = +0.740 V

This is the Eº under standard conditions of 1.0 M Cu2+ and Cd2+. Since the initial concentrations are not standard, we must use the Nernst equation to find the Ecell.

Ecell = Eºcell - RT/nF ln Q and at STP, we have...

Ecell = Eºcell - 0.0592/n log Q

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

Ecell = 0.740 - (0.0296)(log 0.25/1.25) = 0.740 - (0.0296)(-0.699)

Ecell = 0.740 + 0.0207

Ecell = 0.761 V

When voltage drops to 0.729 V, find the [Cu²⁺]. Not 100% sure about the following calculations, but if the math is correct (and it may not be), it kinda makes sense. Hopefully someone else will weigh in with an answer.

Cu²⁺(aq) + Cd(s) ==> Cu(s) + Cd²⁺(aq)

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

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

1.25-x........................................0.25+x.......Equilibrium

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

[Cd²⁺]/[Cu²⁺] = 12.0

0.25+x / 1.25-x = 12

x = 1.13

[Cu²⁺] = 1.25 - 1.13 = 0.12 M