## Electrochemistry intext questions solutions: Class 12 Chemistry Chapter 2 intext questions solutions

Textbook | Ncert |

Class | Class 12 |

Subject | Chemistry |

Chapter | Chapter 2 |

Chapter Name | Electrochemistry Class 12 intext questions solutions |

Category | Intext questions |

Medium | English |

Are you looking for Ncert Chemistry Class 12 intext questions solutions chapter 2 Electrochemistry? Now you can download Electrochemistry intext questions solutions from here.

### Question 2.1:** **How would you determine the standard electrode potential of the system Mg^{2+ }| Mg?

**Solution 2.1:** The standard electrode potential of Mg^{2+ }| Mg can be measured with respect to the standard hydrogen electrode, represented by Pt_{(}_{s}_{)}, H_{2(}_{g}_{)} (1 atm) | H^{+}_{(}_{aq}_{)}(1 M).

A cell, consisting of Mg | MgSO_{4} (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

Mg | Mg^{2+} (aq, 1M)||H^{+}(aq,1M)|H_{2} (g,1 bar),Pt_{(s)}

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

E^{ø} = E^{ø}R – E^{ø}L

Here,E^{ø}R for the standard hydrogen electrode is zero.

∴E^{ø} = 0 – E^{ø}L

= – E^{ø}L

It is observed that the EMF of the cell comes out to be 2.36 V.

Hence, the standard electrode potential for the Mg^{2+ }| Mg system will be E^{Θ} = −2.36 V.

### Question 2.2:** **Can you store copper sulphate solutions in a zinc pot?

**Solution 2.2:** Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution.

Zn + CuSO_{4} → ZnSO_{4} + Cu

Hence, copper sulphate solution cannot be stored in a zinc pot.

### Question 2.3:** **Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

**Solution 2.3:** Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.

Fe^{2+} → Fe^{3+} + e^{-1 } ; E^{ø} = −0.77 V

This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F_{2}, Cl_{2}, and O_{2}.

### Question 2.4: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

**Solution 2.4:** Let’s rework the problem using the correct form of the Nernst equation.

The potential of the hydrogen electrode can be calculated using the Nernst equation:

E = \(E^\circ – \frac{0.0591}{n} \log [H^+]\)

For the hydrogen electrode:

- \( E^\circ = 0 \, \text{V} \) (since it is the standard hydrogen electrode, SHE),
- n = 2 (the number of electrons involved in the reaction),
- \( [H^+] = 10^{-\text{pH}} \).

Since pH = 10, we can write:

\([H^+] = 10^{-10}\)

Now, substitute into the Nernst equation:

E = 0 – \(\frac{0.0591}{1} \log \left( 10^{-10} \right)\)

The logarithm of \( 10^{-10} is -10\), so:

\(E = – 0.0591 \times (-10)\)

\(E = 0.591 \, \text{V}\)

However, because the electrode potential is negative when pH increases, the final answer should be:

E = -0.591V

Thus, the potential of the hydrogen electrode in contact with a solution whose pH is 10 is **-0.591 V**.

### Question 2.5: Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag^{+} (0.002 M) → Ni^{2+ }(0.160 M) + 2Ag(s) Given that E^{ø}cell = 1.05 V

**Solution 2.5:** To calculate the electromotive force (EMF) of the cell using the Nernst equation, we use the following reaction:

\(\text{Ni}(s) + 2\text{Ag}^+(0.002 M) \rightarrow\)\( \text{Ni}^{2+}(0.160 M) + 2\text{Ag}(s)\)

The Nernst equation for a cell is:

\(E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log Q\)

Where:

- \( E^\circ_{\text{cell}} = 1.05 \, \text{V} \) (given).
- n = 2 (since 2 electrons are transferred in the reaction).
- Q is the reaction quotient, given by:

\(Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}\)

Substitute the concentrations:

\(Q = \frac{0.160}{(0.002)^2} \)

\(= \frac{0.160}{0.000004} = 40,000\)

Now, substitute into the Nernst equation:

E_{cell} = \(1.05 – \frac{0.0591}{2} \log (40,000)\)

First, calculate the logarithm of 40,000:

log (40,000) = 4.602

Now, substitute this value:

E_{cell} = \(1.05 – \frac{0.0591}{2} \times 4.602\)

\(E_{\text{cell}} = 1.05 – 0.02955 \times 4.602\)

\(E_{\text{cell}} = 1.05 – 0.136\)

\(E_{\text{cell}} = 0.914 \, \text{V}\)

Thus, the EMF of the cell is **0.914 V**.

### Question 2.6:** **The cell in which the following reactions occurs: 2Fe^{3+} (aq) + 2I^{– }(aq) ⟶ 2Fe^{2+} (aq) +I^{2} (s) has E^{ø}cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

**Solution 2.6:** Here, n = 2, E^{ø}cell = 0.236 T = 298 K

We know that:

### Question 2.7:** **Why does the conductivity of a solution decrease with dilution?

**Solution 2.7:** The conductivity of a solution is linked with the number of ions present per unit volume. With dilution, these decrease and the corresponding conductivity or specific conductance of the solution decreases.

### Question 2.8:** **Suggest a way to determine the Λ^{∘}_{m} value of water.

**Solution 2.8:** Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows:

Λ°m(H_{2}O) = Λ°m(HCl) + Λ°m(NaOH) – Λ°m(NaCl)

Hence, by knowing the Λ°m values of HCl, NaOH, and NaCl, the Λ°m value of water can be determined.

### Question 2.9:** **The molar conductivity of 0.025 mol L^{-1} methanoic acid is 46.1 S cm^{2} mol^{-1}. Calculate its degree of dissociation and dissociation constant Given λ°(H^{+}) = 349.6 S cm^{2} mol^{-1} and λ°(HCOO^{–}) = 54.6 S cm^{2} mol^{-1}.

**Solution 2.9:** To calculate the degree of dissociation \(( \alpha )\) and the dissociation constant \(( K_a )\) of methanoic acid (HCOOH), we follow these steps:

Degree of Dissociation \(( \alpha )\)

The degree of dissociation can be calculated using the formula:

\(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\)

Where:

- \( \Lambda_m \) is the molar conductivity at the given concentration (46.1 S cm² mol⁻¹),
- \( \Lambda_m^\circ \) is the molar conductivity at infinite dilution, calculated as:

\(\Lambda_m^\circ = \lambda^\circ (\text{H}^+) + \lambda^\circ (\text{HCOO}^-)\)

Given:

- \( \lambda^\circ (\text{H}^+) = 349.6 \, \text{S cm}^2 \text{mol}^{-1} \),
- \( \lambda^\circ (\text{HCOO}^-) = 54.6 \, \text{S cm}^2 \text{mol}^{-1} \),

We can calculate \( \Lambda_m^\circ \):

\(\Lambda_m^\circ = 349.6 + 54.6 \)

\(= 404.2 \, \text{S cm}^2 \text{mol}^{-1}\)

Now, calculate \( \alpha \):

\(\alpha = \frac{46.1}{404.2} = 0.114\)

Thus, the degree of dissociation is \( \alpha = 0.114 \) or **11.4%**.

**Step 2: Dissociation Constant \(( K_a )\)**

The dissociation constant for a weak acid is related to the degree of dissociation by the formula:

\(K_a = \frac{c \alpha^2}{1 – \alpha}\)

Where:

- \( c = 0.025 \, \text{mol L}^{-1} \) (the concentration of methanoic acid),
- \( \alpha = 0.114 \).

Now, substitute the values:

\(K_a = \frac{0.025 \times (0.114)^2}{1 – 0.114}\)

First, calculate \( \alpha^2 \):

\(\alpha^2 = (0.114)^2 = 0.013\)

Then calculate the denominator \( 1 – \alpha \):

1 – 0.114 = 0.886

Now, substitute:

\(K_a = \frac{0.025 \times 0.013}{0.886} \)

\(= \frac{0.000325}{0.886} \)

\(= 3.67 \times 10^{-4} \, \text{mol L}^{-1}\)

Thus, the dissociation constant ( K_{a} ) is \( 3.67 \times 10^{-4} \, \text{mol L}^{-1} \).

### Question 2.10:** **If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

**Solution 2.10:** To calculate the number of electrons that flow through a metallic wire when a current of 0.5 A flows for 2 hours, we can use the relationship between electric charge, current, and time, and then convert the charge into the number of electrons.

**Calculate the total charge ( Q )**

The charge ( Q ) is given by the formula:

\(Q = I \times t\)

Where:

- \( I = 0.5 \, \text{A} \) (current),
- \( t = 2 \, \text{hours} \).

First, convert the time from hours to seconds:

t = 2 hours × 3600 seconds/hour = 7200 seconds

Now, calculate the total charge:

Q = 0.5 A × 7200 seconds = 3600 Coulombs

**Calculate the number of electrons**

The charge of a single electron is:

\(e = 1.602 \times 10^{-19} \, \text{C}\)

The number of electrons ( n ) can be calculated by dividing the total charge by the charge of a single electron:

\(n = \frac{Q}{e} = \frac{3600 \, \text{C}}{1.602 \times 10^{-19} \, \text{C/electron}}\)

Now, calculate ( n ):

\(n = \frac{3600}{1.602 \times 10^{-19}} \)

\(\approx 2.247 \times 10^{22} \, \text{electrons}\)

Approximately \( 2.25 \times 10^{22} \)** electrons** would flow through the wire.

### Question 2.11:** **Suggest a list of metals that are extracted electrolytically.

**Solution 2.11:** The highly reactive metals having large -ve E° values, which can themselves act as powerful reducing agents can be extracted electrolytically. The process is known as electrolytic reduction. For details, consult Unit-6. For example, sodium, potassium, calcium, magnesium etc.

### Question 2.12: Consider the reaction: Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{–} ⟶ 2Cr^{3+} + 7H_{2}O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_{2}O_{7}^{2-} ?

**Solution 2.12:** To calculate the quantity of electricity in coulombs needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \), we can follow these steps:

**Identify the number of electrons involved in the reaction**

From the given balanced reaction:

\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow\)\( 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

It shows that **6 electrons** \(( 6e^- )\) are required to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \).

**Use Faraday’s constant**

Faraday’s constant ( F ) is the charge of 1 mole of electrons:

F = 96,485 C/mol

**Calculate the total charge**

Since 6 moles of electrons are involved in the reduction of 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \), the total charge required is:

\(Q = n \times F\)

Where:

- \( n = 6 \, \text{mol of electrons} \),
- \( F = 96,485 \, \text{C/mol} \).

So,

\(Q = 6 \times 96,485 = 578,910 \, \text{C}\)

The quantity of electricity needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) is **578,910 C**.

### Question 2.13: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

**Solution 2.13:** In a lead storage battery, the recharging process involves reversing the chemical reactions that occurred during discharge. During recharging, electrical energy is supplied to the battery, causing the reactants to regenerate from the products formed during discharge. The main reactions involve the lead plates (Pb), lead dioxide plates (PbO₂), and sulfuric acid (H₂SO₄). Here’s the detailed chemistry of the recharging process:

**Discharging Reaction (for reference):**

The reactions during discharge are as follows:

- At the
**negative plate**(lead electrode):

\(\text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^-\) - At the
**positive plate**(lead dioxide electrode):

\(\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \)\(\text{PbSO}_4 + 2\text{H}_2O\)

The overall discharge reaction is:

\(\text{PbO}_2 + \text{Pb} + 2\text{H}_2\text{SO}_4 \rightarrow \)\(2\text{PbSO}_4 + 2\text{H}_2\text{O}\)

Recharging Reaction:

During recharging, electrical energy is applied to reverse the reactions, regenerating lead (Pb), lead dioxide (PbO₂), and sulfuric acid (H₂SO₄). The sulfate ions (SO₄²⁻) leave the lead sulfate (PbSO₄) and combine with the hydrogen ions (H⁺) in the electrolyte, restoring the sulfuric acid.

- At the
**negative plate**(lead electrode):

\(\text{PbSO}_4 + 2e^- \rightarrow \text{Pb} + \text{SO}_4^{2-}\)

Lead sulfate (PbSO₄) on the negative plate is reduced back to lead (Pb). - At the
**positive plate**(lead dioxide electrode):

\(\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \)\(\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^-\)

Lead sulfate (PbSO₄) on the positive plate is oxidized back to lead dioxide (PbO₂).

**Overall Recharging Reaction:**

The overall reaction during recharging is the reverse of the discharge reaction:

\(2\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \)\(\text{PbO}_2 + \text{Pb} + 2\text{H}_2\text{SO}_4\)

**Materials Involved in Recharging:**

**Lead (Pb)**: Re-formed at the negative plate.**Lead dioxide (PbO₂)**: Re-formed at the positive plate.**Lead sulfate (PbSO₄)**: Dissociated from both plates.**Sulfuric acid (H₂SO₄)**: Restored in the electrolyte.**Water (H₂O)**: Produced during discharge and consumed during recharging.

Thus, during the recharging of a lead storage battery, electrical energy is used to convert the lead sulfate (PbSO₄) back into lead (Pb) and lead dioxide (PbO₂), and sulfuric acid (H₂SO₄) is regenerated in the electrolyte.

### Question 2.14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells

**Solution 2.14:** Methane (CH_{4}) and methanol (CH_{3}OH) can also be used as fuels in place of hydrogen in the fuel cells.

### Question 2.15: Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

**Solution 2.15:** In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe_{(s)} → Fe^{2+} (aq) + 2e-

Electrons released at the anodic spot move through the metallic object and go to another spot of the object.

There, in the presence of H^{+} ions, the electrons reduce oxygen. This spot behaves as the cathode. These H^{+} ions come either from H_{2}CO_{3, }which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is given by,

O_{2(g)} + 4H^{+}_{(aq)} + 4e^{– }⟶ 2H_{2}O_{(i)}

The overall reaction is:

2Fe(s)+O_{2}(g)+4H^{+}(aq)⟶2Fe^{2+}(aq)+2H_{2}O(l)+

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide(Fe_{2}O_{3}, x)H_{2}Oi.e., rust.

Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.