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

The mass of Cd is 121.92 g

Step-by-step explanation:

The initial concentrations are:

`nFe=(mass)/(molecular weight) =(100)/(55.845) =1.7906\\nCd=(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-(0.0592)/(2) log([Fe^(2+)] )/([Cd^(2+)] ) \\0.03305=0.038-0.0296log(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

Answer 2

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

Step-by-step 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 = (concentration \ of \ product )/(concentration \ of reactant )`

`= ([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) - (0.059)/(n_e) Q`

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

Substituting for Q

`E = E_(cell) - (0.059)/(n_e) ([Fe^(2+)[Cd]])/([Cd^(2+)][Fe])`

When E = 0.03305 V

`E = E_(cell) - (0.59)/(n_e) (Fe^(2+))/(Cd^(2+))`

Since we are considering the Cd electrode the equation becomes

`E= E_(cell) - (0.059)/(n_e) [(1)/(Cd^(2+)) ]`

Substituting values and making [
`Cd^(2+)`] the subject

`[Cd^(2+)] =\frac{1}{e^{[(0.03305- 0.038)/((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