Question

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer 1

Complete Question

    $Fe^{2+} + 2e^-----> Fe \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = - 0.441 V$

   

    $Cd^{2+} + 2e^- -----> Cd \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = -0.403V$

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer:

The mass is M= 117.37g

Explanation:

The overall reaction  is as follows

             $Cd^{2+} + Fe Fe^{2+} + Cd$

The reaction is this way because the potential  of $Cd^{2+} \ reducing \ to \ Cd$ is higher than the potential  of $Fe^{2+} \ reducing \ to \ Fe$ so the the Fe would be oxidized and $Cd^{2+}$ would be reduced

  At equilibrium the rate constant of the reaction is

                $Q = \frac{concentration \ of \ product }{concentration \ of reactant }$

                      $= \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

The Voltage of the cell $E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}$

     Substituting the given values into the equation

                         $E_{cell} = -0.403 -(-0.441)$

                                 $= 0.038V$

The voltage of the cell at any point can be calculated using the equation

               $E = E_{cell} - \frac{0.059}{n_e} Q$

Where $n_e \ is \ the \ number\ of\ electron$

Substituting for Q

           $E = E_{cell} - \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

 When E = 0.03305 V

            $E = E_{cell} - \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}$

Since we are considering the Cd electrode the equation becomes

            $E= E_{cell} - \frac{0.059}{n_e} [\frac{1}{Cd^{2+}} ]$

Substituting values and making [$Cd^{2+}$]  the subject

          $[Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}$

                      $= 0.8455M$

Given from the question that the volume is 1 Liter

   The number of mole = concentration * volume

                                       = 0.8455 * 1

                                        = 0.8455 moles

At the standard state the concentration of $Cd^{2+}$ is  =1 mole /L

  Hence the amount deposited on the Cd electrode would be

              =  Original amount - The calculated amount

              =   1 - 0.8455

              = 0.1545 moles

The mass deposited is mathematically represented as

             $mass = mole * molar \ mass$

The Molar mass of Cd $= 112.41 g/mol$

          Mass  $= 0.1545 *112.41$

                    $= 17.37g$

Hence the total mass of the electrode is = standard mass + calculated mass

            M= 100+ 17.37

            M= 117.37g

                               

Answer 2

Answer:

The mass of Cd is 121.92 g

Explanation:

The initial concentrations are:

$nFe=\frac{mass}{molecular weight} =\frac{100}{55.845} =1.7906\\nCd=\frac{100}{112.411} =0.88959$

Initially:

[Fe2+] = 1.7906[Cd2+] = 0.88959

After the reaction:

[Fe2+] = 1.7906 + x

[Cd2+] = 0.88959 - x

Eo = Ered - Eoxidation = (-0.403 - 0.441) = 0.038

The Nernst equation is:

$E=0.038-\frac{0.0592}{2} log\frac{[Fe^{2+}] }{[Cd^{2+}] } \\0.03305=0.038-0.0296log\frac{1.7906+x}{0.88959-x}$

Solving for x:

x = -0.195

[Fe2+] = 1.7906 -0.195 = 1.5956

[Cd2+] = 0.88959 + 0.066 = 1.08459

Mass of Cd2+ = 1.08459 * 112.411 = 121.92 g