Quiz 3 - Electrochemistry | Electrochemistry

The anode is Cd, as it has the lower reduction potential, while Ag is the cathode. The standard cell voltage is calculated as:
E^o_cell = E^o_cathode - E^o_anode = 0.80 V - (-0.40 V) = 1.20 V

In the electrolysis of aluminum oxide (Al₂O₃​), the reaction involves breaking down Al₂O₃​ into aluminum and oxygen gas:
Al₂O₃ → 2 Al (s) + $\frac{3}{2}$ O₂ (g)

From the reaction, 1 mole of Al₂O₃​ produces 2 moles of aluminum and $\frac{3}{2}$​ moles of oxygen gas.

Thus, the ratio of moles of aluminum to moles of oxygen gas is: 2:$\frac{3}{2}$ = 4:3

The formula for calculating the standard Gibbs free energy change (ΔG^o) for a reaction is:

Given that the balanced reaction involves 6 electrons (from 3 Cu²⁺ ions each accepting 2 electrons), we have:

ΔG^o (in J.mol^-1) = − 6 x 96,500 x 0.43

ΔG^o (in kJ.mol^-1) = - $\frac{6 \times 96500 \times 0.43}{1000}$

The Nernst equation is used to calculate the cell potential under non-standard conditions:
 

E_cell = E^o_cell - $\frac{0.0592}{\mathrm{n}}$ log $\frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{{\left[{\mathrm{Cu}}^{2+}\right]}^3}$ = 0.43 - $\frac{0.0592}{6}$ log $\frac{0.{010}^2}{1^3}$ = 0.43 + 0.039 = 0.47 V

During the electrolysis of a molten salt, the cation (positively charged ion) is attracted to the cathode (negative electrode). At the cathode, the cation gains electrons and undergoes reduction.

• Reduction half-reaction (as given for Tl⁺): Tl⁺ (aq) + e⁻ → Tl (s), E^o = −0.336 V

• Oxidation half-reaction (reverse of Ga³⁺ reduction): Ga (s) → Ga³⁺ (aq) + 3 e⁻, E^o = +0.529 V

The cell potential E^o_cell​ is given by: E^o_cell = E^o_reduction + E^o_oxidation = -0.336 + 0.529 = 0.193 V

• Reduction (Cu²⁺ to Cu): Cu²⁺ (aq) + 2 e⁻ → Cu(s), E^o = +0.34 V

• Oxidation (Al to Al³⁺): Al (s) → Al³⁺ (aq) + 3 e⁻, E^o = −(−1.66) = +1.66 V

The cell potential is: E^o_cell = E^o_reduction +E^o_oxidation = 0.34 + 1.66 = 2.00 V

The Nernst equation requires n, the total number of moles of electrons transferred in the balanced reaction. For:

2 Al (s) + 3 Cu²⁺ (aq) → 2 Al³⁺ (aq) + 3 Cu (s)

• Each Al atom loses 3 electrons (oxidation: 2 x 3 = 6 electrons total).
• Each Cu²⁺ ion gains 2 electrons (reduction: 3 x 2 = 6 electrons total).

Thus, n = 6

According to the Farday's first law:

The reduction of Cr₂O₇²⁻​ to Cr²⁺ involves the following reaction: Cr₂O₇²⁻ + 14 H⁺ + 8 e⁻ → 2 Cr²⁺ + 7 H₂O
Each Cr₂O₇²⁻​ molecule requires 8 electrons to fully reduce to 2 Cr²⁺ ⇒ n = 8

Calculate moles of Cr₂O₇²⁻:
n_Cr2O7^2- = [Cr₂O₇²⁻] x V = 0.115 x 0.150 = 1.725 x 10^-2 mol

Thus, Q = 6 x 1.725 x 10^-2 x F = 0.138 F

According to the Faraday's first law:

The total charge (Q) passed is given by: Q = I x t = 10.0 x 30.6 = 306 C

The reduction reaction for Cu²⁺ to Cu is: Cu²⁺ + 2 e⁻ → Cu(s) ⇒ Each mole of Cu²⁺ requires 2 moles of electrons ⇒ n = 2

Therefore, m_Cu = $\frac{\mathrm{Q}}{\mathrm{F}}$ x $\frac{{\mathrm{M}}_{\mathrm{Cu}}}{n}$ = $\frac{306}{96485}$ x $\frac{63.546}{2}$ = 0.101 g