Electrochemical Cell: Definition, Description, Applications and Types

Ans. Electrochemical cells are devices that have the capacity to generate electrical energy from chemical reactions or cause chemical reactions by using electrical energy.

Ans. An electrolytic cell is a type of electrochemical cell. It converts electrical energy into chemical energy and the electrons flow from the anode to the cathode with the help of external sources such as batteries. In the electrolyte solution only ions flow, electrons do not.

Ans. Standard conditions are that which takes place at 298K temperature, 1atm pressure, and have a molarity of 1.0M for both anode and cathode solutions. 

Ans. The main components of electrochemical cells are:

• The anode
• The cathode
• External sources
• Salt bridge or porous barrier.

Ans. The disadvantages of electrochemical cells are:

• It can not be recycled.
• It has a short shelf-life.
• Unstable voltage and less power.

Answer:  Molar conductivity: Molar conductivity of a solution can be defined as the conductivity of an electrolyte solution divided by molar concentration of an electrolyte. It is used to measure the efficiency of an electrolyte which is used to conduct the electricity in an electrolyte. It is represented by Λm (lambda).

Λm =KAl ( Since, l = 1 and A = V)

So, Λm = KV 

Unit = S cm² mol^-1

In case of strong electrolytes there is a less increase in conductivity with dilution because a strong electrolyte gets completely scattered in solution and the number of ions remains constant. In case of weak electrolytes there will be an increase in conductance with a decrease in concentration as the number of ions in the solution increases.

The graph between Λm and concentration also shows the same.

Answer: 

Reaction at the Voltaic Cell-

\(\begin{aligned} &\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{+2}+2 \mathrm{Ag}\\ &\quad \quad (0.001\ \mathrm{M}) \quad(0.10)\\ &\mathrm{Cu} / \mathrm{Cu}^{+2}(0.10) \| \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}\\ &E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}\\ &E_{\text {cell }}=0.46-\frac{0.059}{2} \log \frac{0.1}{(0.001)^{2}}\\ &=0.46-\frac{0.059}{2} \log 10^{5}\\ &=0.46-\frac{0.059}{2} \times 5 \log 10\\ &=0.46-0.1475\\ &=0.3125 \end{aligned}\)

Answer:

(a) The relationship between the cell constant of a cell (G*), resistance of the solution in the cell (R) and conductivity (K) is given by

\(K = \frac{\text{Cell constant}}{\mathrm{R}} = \frac{\mathrm{G^*}}{\mathrm{R}}\)

The relationship between molar conductivity (Λm) and conductivity of the solution (K) is given by

Λm =K / C

(b) The cell may be represented as

Al | Al³⁺ || Ni²⁺| Ni

E⁰_cell = E⁰_R – E⁰_L

E⁰_cell = (-0-25) – (-1.66)

∴ E⁰ = -0.25 + 1.66 = 1.41 V

The reaction takes place at anode and cathode in the following ways :

At anode (oxidation) :

Cu(s) → Cu²⁺(aq) + 2e^–

At cathode (reduction) :

Cu(s) + 2Ag²⁺(aq) → Cu²⁺(aq) + 2Ag(s)

The complete cell reaction can be given as-

Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)

∵  E⁰_cell = E⁰_cathode – E⁰_anode

or E⁰_cell = + 0.80 – (+ 0.34)

or E⁰_cell = 0.80 – 0.34 = 0.46 V

Using Nernst equation

\(\begin{array}{ll} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{2+}(\mathrm{aq})\right]}{\left[\mathrm{Ag}^{+}(\mathrm{aq})\right]^{2}} \\ & 0.422=0.46-\frac{0.059}{2} \log \frac{(0.1)}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.422-0.46=-\frac{0.059}{2} \log \frac{10^{-1}}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & -0.038=-0.0295\left[\log 10^{-1}-\log \left[\mathrm{Ag}^{+}\right]^{2}\right. \\ & -0.038=-0.0295\left[-1-2 \log \left[\mathrm{Ag}^{+}\right]\right] \\ & -0.038=0.0295+0.059 \log \left[\mathrm{Ag}^{+}\right] \\ \text { or } & -0.059 \log [\mathrm{Ag}^{+}]=0.038+0.0295 \\ \text { or } & -0.059 \log [\mathrm{Ag}^{+}]=0.0675 \\ \text { or } & -\log [\mathrm{Ag}^{+}]=\frac{0.0675}{-0.059} \\ \text { or } & \log \left[\mathrm{Ag}{ }^{+}\right]=1.14407 \\ \text { or } & {\left[\mathrm{Ag}{ }^{+}\right]=\mathrm{Antilog}\ 1.14407} \\ \therefore & {\left[\mathrm{Ag}{ }^{+}\right]=13.93\ \mathrm{M}} \end{array}\)

Answer:

(a) At anode:

\(\mathrm{Pb} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{PbSO_4} + 2e^- (\text{oxidation}) \)

At cathode:

\(\mathrm{PbO_2} + \mathrm{SO}_4^{2-} + 4\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{PbSO_4} + 2\mathrm{H_2O} (\text{reduction})\)

Overall reaction:

\(\mathrm{Pb} + \mathrm{PbO_2} + 4\mathrm{H}^{+} + 2\mathrm{SO}_4^{2-} \rightarrow 2\mathrm{PbSO_4} + 2\mathrm{H_2O} \)

Answer: (a) M = 0.20 mol L^-1, κ = 2.48 × 10^-2 S cm^-1, Λ_m = ?, α = ?

\(\begin{aligned} &\lambda_{(\mathrm{K^+})}^{\circ} = 73.5\ \mathrm{S\ cm^2 mol^{-1}},\quad \lambda_{(\mathrm{Cl^-})}^{\circ} = 76.5\ \mathrm{S\ cm^2 mol^{-1}}\\ &\Lambda_{(\mathrm{KCl})}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}=73.5+76.5=150.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\\ &\text { Now, } \quad \Lambda_{m}=\frac{1000 \kappa}{M}=\frac{1000 \times 2.48 \times 10^{-2}}{0.20}\\ &\Rightarrow \quad \Lambda_{\mathrm{m}}=\frac{248}{2}=124 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\\ &\therefore \quad \alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}=\frac{124}{150.0}=8.26 \times 10^{-1}\\ &\Rightarrow \quad \alpha=8.26 \times 10^{-1} \times 10^{2}=82.6 \% \end{aligned}\)

(b) The Mercury cell is a primary cell because it is not rechargeable. It has more efficiency than the dry cell and its voltage remains constant over a large period of time.

Answer: 

(a) Lead storage battery: It is a secondary cell. It cosists of a lead anode and a grid of lead packed with lead dioxide as cathode. A solution of sulphuric acid (38% by mass) is used as an electrolyte.

The cell reactions when the battery is in use are given below:

Anode: Pb(s) + SO₄^2-(aq) → PbSO₄(s) + 2e^-

Cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^- → PbSO₄(s) + 2H₂O(l)

The overall cell reaction consisting of cathode and anode reactions is:

Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

On recharging the battery the reaction is reversed.

(b) \(\begin{align} \mathrm{Zn}(s) &\longrightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-} \\ \mathrm{Ag}^{+}(a q)+2 e^{-} & \longrightarrow 2 \mathrm{Ag}(s) \\ \hline \mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) & \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s) \\ \hline \end{align}\)

\(\begin{aligned} \mathbf{E}_{\text {cell }}^{\circ} &=\left(\mathbf{E}_{\substack{\text { SRP } \\ \text { callode }}}^{\circ}-\mathbf{E}_{\text {anode }}^{\circ}\right)\\ &=+0.80 \mathrm{~V}-(-0.76 \mathrm{~V})=+1.56 \mathrm{~V} \end{aligned}\)

\(\begin{aligned} \Delta_{r} \mathrm{G}^{\circ}&=-n\ \mathrm{E}^{\circ} \mathrm{F}\\ \Delta_{r} \mathrm{G}^{\circ}&=-2 \times 1.56 \mathrm{~V} \times 96500=-301080 \mathrm{~J}\ \mathrm{mol}^{-1}\\ &=-301.080 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)

Answer: 

κ = 1.29 × 10^-2 ohm^-1 cm^-1

\(\begin{aligned} \quad \quad \quad \kappa &=\frac{1}{R} \times \frac{1}{A} \Rightarrow \frac{l}{A}=\kappa \times R=1.29 \times 10^{-2} \times 85 \\ &=109.65 \times 10^{-2}=1.0965 \mathrm{~cm}^{-1} \end{aligned}\)

\(\begin{aligned} \Lambda_{m} &=\frac{1000 \kappa}{M}=\frac{1000}{M} \times \frac{1}{R} \times \frac{l}{A} \\ \Rightarrow \quad \Lambda_{m} &=\frac{1000 \times 1 \times 1.0965}{0.052 \times 96}=\frac{1096.50}{4.992}=219.65 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \end{aligned}\)

Answer: 

\(\begin{align} \mathrm{A} &\longrightarrow \mathrm{A}^{2+}(aq) + 2e^- \\ \mathrm{B}^{2+}(aq) + 2e^- & \longrightarrow \mathrm{B} \\ \hline \mathrm{A} + \mathrm{B}^{2+} &\longrightarrow \mathrm{A}^{2+} + \mathrm{B} \\ \hline \end{align}\)\(\begin{align} &\\ &\\ &\\ \quad \quad \quad \therefore n=2 \\ \end{align}\)

\(\begin{aligned} \mathrm{E}_{\text {cell }} &=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{A}^{2+}\right]}{\left[\mathrm{B}^{2+}\right]} \\ 2.6805 \mathrm{~V} &=\mathrm{E}_{\text {Cell }}^{\circ}-\frac{0.0591}{2} \log \frac{10^{-4}}{10^{-3}}=\mathrm{E}_{\text {Cell }}^{\circ}+0.0295 \log 10 \\ \mathrm{E}_{\text {cell }}^{\circ} &=2.6805-0.0295=2.651 \mathrm{~V} \end{aligned}\)

Answer: 

(a) \(\begin{align} \mathrm{Mg}(s) &\longrightarrow \mathrm{Mg}^{2+}(aq) + 2e^- \\ \mathrm{Cu}^{2+}(aq) + 2e^- & \longrightarrow \mathrm{Cu(s)} \\ \hline \mathrm{Mg}(s) + \mathrm{Cu}^{2+}(aq) &\longrightarrow \mathrm{Mg}^{2+}(aq) + \mathrm{Cu}(s) \\ \hline n&=2 \end{align}\)

\(\begin{aligned} &\Delta_{r} G^{\circ}=-n\ E^{\circ} \mathrm{F}=-2 \times 2.71 \mathrm{~V} \times 96500 \mathrm{C}=-523030 \mathrm{~J} \mathrm{~mol}^{-1} \\ &\Delta_{r} G^{\circ}=-523.030 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)

(b) H₂—O₂ Fuel Cell was used in the Apollo space programme for providing electrical power.

Answer: (i) Molar conductivity: Molar Conductivity can be defined as the conductance of a solution having 1 mole of electrolyte which is placed in a cell having electrodes unit distance apart with sufficient area of cross-section to carry the electrolyte.

(ii) Secondary battery: The batteries that can be recharged are called secondary batteries, e.g. lead storage battery, Nickel-cadmium battery.

Answer: 

Zn(s) → Zn²⁺(aq) + 2e^-             At anode

Cu²⁺(aq) + 2e^- → Cu(s)            At cathode

n = 2

\(\begin{aligned} \Delta \mathrm{G}^{\circ} &=-n\ \mathrm{E}^{\circ} \mathrm{F} \\ &=\frac{-2 \times 1.10 \mathrm{~V} \times 96500 \mathrm{C}}{1000}=-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad[\mathrm{CV}=\mathrm{J}] \end{aligned}\)