C12 C2 Electrochemistry Short Question Answers

Chapter 2: Electrochemistry

Short Question Answers

i. The oxidation potential of Zn is +0.76V and its reduction potential is -0.76V

The oxidation and reduction potentials for the same half-reaction are equal in magnitude but opposite in sign. Oxidation is the reverse process of reduction. Therefore, the potential for one process will be the negative of the potential for the other. This can be represented by the following equations:

$$E_{\text{oxidation}} = -E_{\text{reduction}}$$

For zinc (Zn), the oxidation half-reaction is:

$$Zn \rightarrow Zn^{2+} + 2e^- \quad E^\circ = +0.76V$$

The reduction half-reaction is:

$$Zn^{2+} + 2e^- \rightarrow Zn \quad E^\circ = -0.76V$$

ii. A salt bridge maintains the electrical neutrality in the cell

A salt bridge connects the two half-cells of a galvanic cell and contains an electrolyte solution. As oxidation occurs at the anode, positive ions accumulate in that half-cell. Simultaneously, as reduction occurs at the cathode, negative ions accumulate in that half-cell. The salt bridge allows ions to migrate between the two half-cells to balance these charges, preventing a buildup that would stop the flow of electrons and bring the reaction to a halt.

iii. Na and K can displace hydrogen from acids but Cu and Pt cannot

This is determined by the **activity series** of metals. Metals higher up in the series are more reactive and have a greater tendency to be oxidized (lose electrons) than those below them. Sodium (Na) and Potassium (K) are highly reactive alkali metals, located high in the activity series, meaning their standard reduction potentials are very negative. They are therefore able to reduce the hydrogen ions ($$H^+$$) in acids to hydrogen gas ($$H_2$$).

In contrast, Copper (Cu) and Platinum (Pt) are less reactive and are located below hydrogen in the activity series, with positive standard reduction potentials. They have a lower tendency to be oxidized and therefore cannot displace hydrogen from acids.

$$Na \rightarrow Na^+ + e^-$$

$$K \rightarrow K^+ + e^-$$

$$2H^+ + 2e^- \rightarrow H_2$$

$$Cu \rightarrow Cu^{2+} + 2e^-$$

iv. Define oxidation in terms of electron transfer

Oxidation is the **loss of electrons** by a substance during a chemical reaction. A substance that is oxidized loses one or more electrons, and its oxidation number increases.

v. What is the oxidation number of oxygen in $$H_2O_2$$?

The oxidation number of hydrogen (H) is typically +1. Since the molecule $$H_2O_2$$ is neutral, the sum of the oxidation numbers must be zero. Let 'x' be the oxidation number of oxygen (O).

$$2(+1) + 2(x) = 0$$

$$2 + 2x = 0$$

$$2x = -2$$

$$x = -1$$

Therefore, the oxidation number of oxygen in hydrogen peroxide ($$H_2O_2$$) is **-1**.

vi. Identify the reducing agent in the reaction $$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$$

A **reducing agent** is the substance that gets oxidized (loses electrons) and causes the reduction of another substance. In the given reaction:

• $$Zn$$ goes from an oxidation number of 0 to +2. It loses electrons, so it is oxidized.
• $$Cu^{2+}$$ goes from an oxidation number of +2 to 0. It gains electrons, so it is reduced.

Since **Zn** is oxidized, it is the reducing agent.

vii. State the purpose of the Winkler method

The Winkler method is a titration technique used to **determine the concentration of dissolved oxygen (DO)** in water samples. It is a key procedure in environmental and water quality monitoring.

viii. Explain what happens at the anode during the electrolysis of aqueous sodium chloride

In the electrolysis of aqueous sodium chloride ($$NaCl$$ solution), the anode is the positive electrode. During electrolysis, **oxidation** occurs at the anode. The species present are $$Cl^-$$ ions from $$NaCl$$ and $$H_2O$$. Oxidation potentials are compared to see which reaction is more likely to occur:

• Oxidation of chloride ions: $$2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \quad E^\circ = -1.36V$$
• Oxidation of water: $$2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \quad E^\circ = -1.23V$$

Based on the standard potentials, the oxidation of water is more favorable. However, due to the high **overpotential** of oxygen gas formation, the oxidation of chloride ions to form chlorine gas ($$Cl_2$$) is the reaction that predominantly occurs at the anode.

ix. What does a positive standard electrode potential indicate about a substance's tendency to gain electrons?

A positive standard electrode potential ($$E^\circ > 0$$) indicates a **high tendency for a substance to gain electrons** and undergo reduction. The more positive the potential, the stronger the oxidizing agent it is.

x. Calculate the oxidation number of chromium in $$K_2Cr_2O_7$$

The compound $$K_2Cr_2O_7$$ is neutral, so the sum of the oxidation numbers must be zero. The oxidation number of Potassium (K) is +1 (alkali metal), and oxygen (O) is typically -2. Let 'x' be the oxidation number of Chromium (Cr).

$$2(K) + 2(Cr) + 7(O) = 0$$

$$2(+1) + 2(x) + 7(-2) = 0$$

$$2 + 2x - 14 = 0$$

$$2x - 12 = 0$$

$$2x = 12$$

$$x = +6$$

The oxidation number of chromium in $$K_2Cr_2O_7$$ is **+6**.

xiv. Calculation of mass of Al and volume of O₂

First, let's calculate the total charge passed. The current (I) is 15 A, and the time (t) is 10 hours. We need to convert the time to seconds:

$$t = 10 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 36000 \text{ s}$$

The total charge (Q) is given by the formula $$Q = I \times t$$

$$Q = 15 \text{ A} \times 36000 \text{ s} = 540000 \text{ C}$$

Now, let's find the number of moles of electrons (nₑ⁻) using the Faraday constant ($$F \approx 96485 \text{ C/mol e}^-$$):

$$n_{e^-} = \frac{Q}{F} = \frac{540000 \text{ C}}{96485 \text{ C/mol e}^-} \approx 5.60 \text{ mol e}^-$$

The half-reactions are:

• **At the cathode (Reduction):** $$Al^{3+} + 3e^- \rightarrow Al(s)$$
• **At the anode (Oxidation):** $$2O^{2-} \rightarrow O_2(g) + 4e^-$$

From the half-reactions, we can see the molar ratios:

• 1 mol Al requires 3 mol e⁻
• 1 mol O₂ requires 4 mol e⁻

Using these ratios, let's find the moles of Al and O₂:

• $$n_{Al} = 5.60 \text{ mol e}^- \times \frac{1 \text{ mol Al}}{3 \text{ mol e}^-} \approx 1.867 \text{ mol Al}$$
• $$n_{O_2} = 5.60 \text{ mol e}^- \times \frac{1 \text{ mol } O_2}{4 \text{ mol e}^-} = 1.4 \text{ mol } O_2$$

Now, let's calculate the mass of Al and the volume of O₂.

Mass of Al:

Molar mass of Al is approximately 26.98 g/mol.

$$\text{Mass of Al} = n_{Al} \times \text{Molar mass}_{Al}$$

$$\text{Mass of Al} = 1.867 \text{ mol} \times 26.98 \text{ g/mol} \approx 50.37 \text{ g}$$

Volume of O₂:

We use the ideal gas law, $$PV = nRT$$. We are given standard conditions (1 atm and 25°C). The gas constant R for these units is $$0.0821 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K})$$.

Temperature in Kelvin: $$T = 25 + 273.15 = 298.15 \text{ K}$$

$$V = \frac{nRT}{P} = \frac{(1.4 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))(298.15 \text{ K})}{1 \text{ atm}}$$

$$V \approx 34.3 \text{ L}$$

The answer provided is 34.20 dm³, and 1 L = 1 dm³, so the volume is approximately 34.3 dm³.

Answer: Mass of Al is approximately 50.37 g and the volume of O₂ is approximately 34.3 dm³.

xv. Comparison of Mass Deposited

The mass of metal deposited is directly proportional to its molar mass and inversely proportional to the number of electrons required for its reduction. The formula is:

$$\text{Mass} = \frac{I \times t \times \text{Molar Mass}}{F \times \text{Charge}}$$

Since the current (I), time (t), and Faraday constant (F) are the same for both cases, we can compare the ratio of Molar Mass to the charge for each metal.

• For NaCl, the reaction is $$Na^+ + e^- \rightarrow Na$$. The charge on Na is +1.

  Molar Mass of Na is approx. 23 g/mol. Ratio: $$\frac{23}{1} = 23$$

• For $$CaCl_2$$, the reaction is $$Ca^{2+} + 2e^- \rightarrow Ca$$. The charge on Ca is +2.

  Molar Mass of Ca is approx. 40 g/mol. Ratio: $$\frac{40}{2} = 20$$

Since the ratio for **Na is greater (23 > 20)**, **NaCl will give more mass of metal.**

xvi. Electroplating Time Calculation

We need to find the time (t) to deposit 75 g of copper with a 5 A current. The reaction is: $$Cu^{2+} + 2e^- \rightarrow Cu(s)$$.

First, find the moles of Cu to be deposited:

Molar mass of Cu is approx. 63.55 g/mol.

$$n_{Cu} = \frac{75 \text{ g}}{63.55 \text{ g/mol}} \approx 1.18 \text{ mol Cu}$$

From the reaction, 1 mol of Cu requires 2 mol of electrons. So, moles of electrons needed is:

$$n_{e^-} = 1.18 \text{ mol Cu} \times \frac{2 \text{ mol e}^-}{1 \text{ mol Cu}} = 2.36 \text{ mol e}^-$$

Now, find the total charge required:

$$Q = n_{e^-} \times F = 2.36 \text{ mol} \times 96485 \text{ C/mol} \approx 227600 \text{ C}$$

Finally, calculate the time using $$t = Q/I$$:

$$t = \frac{227600 \text{ C}}{5 \text{ A}} = 45520 \text{ s}$$

Convert seconds to hours:

$$\text{Hours} = \frac{45520 \text{ s}}{3600 \text{ s/hr}} \approx 12.64 \text{ hours}$$

Answer: It would take approximately 12.64 hours.

xvii. Differentiation

(a) Galvanic vs. Electrolytic Cell

• **Galvanic Cell (Voltaic Cell):** Converts **chemical energy into electrical energy**. It is a **spontaneous** reaction ($$ \Delta G < 0 $$) that produces a voltage. Examples include batteries.
• **Electrolytic Cell:** Converts **electrical energy into chemical energy**. It requires an external power source to drive a **non-spontaneous** reaction ($$ \Delta G > 0 $$). Example: electrolysis of water.

(b) Oxidation Half-Reaction vs. Reduction Half-Reaction

• **Oxidation Half-Reaction:** Describes the process of a substance **losing electrons**. The oxidation number of the element **increases**. It occurs at the **anode**.
• **Reduction Half-Reaction:** Describes the process of a substance **gaining electrons**. The oxidation number of the element **decreases**. It occurs at the **cathode**.

xviii. Electroplating Gold Calculation

We need to find the mass of gold deposited. The half-reaction is given: $$[Au(CN)_4]^- + 3e^- \rightarrow Au(s) + 4CN^-$$

First, calculate the total charge (Q) passed. Current (I) = 5.0 A, time (t) = 30 minutes. Convert time to seconds:

$$t = 30 \text{ min} \times 60 \text{ s/min} = 1800 \text{ s}$$

$$Q = I \times t = 5.0 \text{ A} \times 1800 \text{ s} = 9000 \text{ C}$$

Next, find the moles of electrons ($$n_{e^-}$$):

$$n_{e^-} = \frac{Q}{F} = \frac{9000 \text{ C}}{96485 \text{ C/mol e}^-} \approx 0.0933 \text{ mol e}^-$$

From the half-reaction, 1 mol of Au is deposited for every 3 mol of electrons. Moles of Au deposited is:

$$n_{Au} = 0.0933 \text{ mol e}^- \times \frac{1 \text{ mol Au}}{3 \text{ mol e}^-} \approx 0.0311 \text{ mol Au}$$

Finally, calculate the mass of Au using its molar mass (approx. 196.97 g/mol):

$$\text{Mass of Au} = n_{Au} \times \text{Molar mass}_{Au}$$

$$\text{Mass of Au} = 0.0311 \text{ mol} \times 196.97 \text{ g/mol} \approx 6.12 \text{ g}$$

Answer: The mass of gold deposited is approximately 6.12 g.

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