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 of Cd is 121.92 g
Explanation:
The initial concentrations are:
[tex]nFe=\frac{mass}{molecular weight} =\frac{100}{55.845} =1.7906\\nCd=\frac{100}{112.411} =0.88959[/tex]
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:
[tex]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}[/tex]
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
Complete Question
[tex]Fe^{2+} + 2e^—–> Fe \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = – 0.441 V[/tex]
[tex]Cd^{2+} + 2e^- —–> Cd \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = -0.403V[/tex]
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
[tex]Cd^{2+} + Fe <========> Fe^{2+} + Cd[/tex]
The reaction is this way because the potential of [tex]Cd^{2+} \ reducing \ to \ Cd[/tex] is higher than the potential of [tex]Fe^{2+} \ reducing \ to \ Fe[/tex] so the the Fe would be oxidized and [tex]Cd^{2+}[/tex] would be reduced
At equilibrium the rate constant of the reaction is
[tex]Q = \frac{concentration \ of \ product }{concentration \ of reactant }[/tex]
[tex]= \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}[/tex]
The Voltage of the cell [tex]E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}[/tex]
Substituting the given values into the equation
[tex]E_{cell} = -0.403 -(-0.441)[/tex]
[tex]= 0.038V[/tex]
The voltage of the cell at any point can be calculated using the equation
[tex]E = E_{cell} – \frac{0.059}{n_e} Q[/tex]
Where [tex]n_e \ is \ the \ number\ of\ electron[/tex]
Substituting for Q
[tex]E = E_{cell} – \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}[/tex]
When E = 0.03305 V
[tex]E = E_{cell} – \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}[/tex]
Since we are considering the Cd electrode the equation becomes
[tex]E= E_{cell} – \frac{0.059}{n_e} [\frac{1}{Cd^{2+}} ][/tex]
Substituting values and making [[tex]Cd^{2+}[/tex]] the subject
[tex][Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}[/tex]
[tex]= 0.8455M[/tex]
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 [tex]Cd^{2+}[/tex] 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
[tex]mass = mole * molar \ mass[/tex]
The Molar mass of Cd [tex]= 112.41 g/mol[/tex]
Mass [tex]= 0.1545 *112.41[/tex]
[tex]= 17.37g[/tex]
Hence the total mass of the electrode is = standard mass + calculated mass
M= 100+ 17.37
M= 117.37g