Electrochemistry - Class 12 Medical Chemistry - Extra Questions

At anode : $$Ag(s)\, +\, Cl^{\ominus}\, \rightarrow\, AgCl(s)\, +\, e^{-}$$

At cathode : $$Ag^{\oplus}(aq)\, +\, e^{-}\, \rightarrow\, Ag(s)$$

$$\Delta_{f}G^{\ominus}\, =\, -\, nFE^{\ominus}_{cell}\, =\, -\, 1\, \times\, 96500\, \times\, E^{\ominus}_{cell}$$

$$-\, 57\, \times\, 10^{3}\, J\, mol^{-1}\, =\, -\,96500 E^{\ominus}_{cell}$$

$$\therefore\, E^{\ominus}_{cell}\, =\, \displaystyle\, \dfrac{-\, 57\, \times\, 10^{3}}{-\, 96500}\, =\, 0.59\, V$$

$$\Delta G =-nFE$$
$$= -2\times 96500\times 2.2$$  
= $$-212300\ J$$
= $$-212.3\ kJ$$

That photo frame has covered with the layer of rust.

Rusting is the process in which iron turns into iron oxide. These are formed by the redox reaction of iron and oxygen in the presence of water or air moisture. The process is a type of corrosion that occurs easily under natural conditions.

(a) The effect of temperature on the rate of a reaction:
(i) With increase in temperature, intermolecular collisions  will be more violent. This will result in higher molecular velocities and less time between collisions. The collision frequency and the number of effective collisions will increase which will increase the rate for the reaction.  For every 10 degree rise in temperature, the rate of reaction is roughly doubled.
(ii) The effect of catalyst on the rate of a reaction:
A catalyst provides an alternate pathway of lower activation energy and increases the rate of the reaction.

(b) Faraday's first law of electrolysis:
The amount of the substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.
Thus mass of the substance produced
$$\displaystyle = \dfrac {\text {I(A) } \times \text {t(s)}}{\text {96500 C/mol e}^-} \times \text {mole ratio} \times \text {molar mass}$$
Here I(A) is current in ampere and t(s) is time in second. 
Number of moles of electrons passed
$$\displaystyle = \dfrac {I(A) \times t(s)}{96500}= \dfrac {1.5 \times 10 \times 60}{96500}= 0.009326$$ moles of electrons
$$\displaystyle Cu^{2+}+2e^- \rightarrow Cu$$
The number of moles of copper deposited $$\displaystyle = \dfrac {0.009326}{2}=0.004663$$ moles.
The molar mass of Cu is 63.5 g/mol.
The mass of Cu deposited $$\displaystyle =0.004663 \: mol \times 63.5 \: g/mol = 0.296 \: g$$

Faraday's first law of electrolysis:
The amount of the substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

Thus mass of the substance produced
$$\displaystyle = \dfrac {\text {I(A) } \times \text {t(s)}}{\text {96500 C/mol e}^-} \times \text {mole ratio} \times \text {molar mass}$$
Here I(A) is current in ampere and t(s) is time in second.

Faraday's second law of electrolysis:
When the same amount of electricity is passed through different cells containing different electrolytes and arranged in series, the amounts of substances oxidized or reduced at the respective electrodes are directly proportional to their chemical equivalent masses.

$$\displaystyle \dfrac {  \text { moles of A produced}}{  \text { moles of B produced}} = \dfrac {  \text { mole ratio of A half reaction }}{  \text { mole ratio of B half reaction }}$$

$$\displaystyle \kappa = \dfrac {1}{R}\dfrac {l}{a}$$
Here $$\displaystyle \kappa$$ is conductivity, R is resistance and $$\displaystyle \dfrac {l}{a}$$ is the cell constant.

For $$0.15\ M$$ solution
$$\displaystyle \kappa = \dfrac {1}{R}\dfrac {l}{a}$$
$$\displaystyle 1.5 s/m = \dfrac {1}{50 \Omega}\dfrac {l}{a}$$
$$\displaystyle  \dfrac {l}{a}=75 /m$$

For $$0.02\ M$$ solution
$$\displaystyle \kappa = \dfrac {1}{R}\dfrac {l}{a}$$

Electrolytic cell	Galvanic cell
It requires a source of external energy	It is a source of energy
It converts electrical energy into chemical energy	Converts chemical energy into electrical energy
Has cathod as the negative electrode	Has cathode as positive electrode
Has anode as the positive electrode	Has anode as negative electrode

Faraday's First law of electrolysis:
The amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

Mathematical form :
(i) Moles of the substance produced $$\displaystyle = \dfrac {I(A) \times t(s)}{96500 (C/mol e^-)} \times$$ mole ratio from stoichiometry

(ii) Mass of the substance produced $$\displaystyle = \dfrac {I(A) \times t(s)}{96500 (C/mol e^-)} \times$$   mole ratio  $$\displaystyle  \times$$ molar mass of the substance

2 faraday are required to deposit one mole of zinc. This is because 2 moles of electrons are required to deposit one mole of zinc.This corresponds to $$\displaystyle 2 \times 96500=193000$$ coulombs.

Standard electrode potential : The standard electrode potential, abbreviated as $$E$$ , is the measure of potential of a reaction that occurs at the electrode when all the substances involved in the reaction are in their standard states that is solutions are at $$1 M$$ concentrations, gases at $$1$$ atm pressure and solids and liquids are in pure form with all at $$25 C$$ .

• Iron in presence of moist air is oxidized to give rust. This process is known as corrosion.
• Hydrated Iron (III) oxide is known as rust.
• Chemical formula is $$Fe_2O_3.xH_2O$$ .

$$E_{cell}^{\circ} = 0.03\ volt, \triangle G^{\circ} = -nFE^{\circ}_{cell}$$
$$-2895\ J$$ .

$$K=\cfrac{1}{\rho}$$

Galvanic cell or voltaic cell is an electrochemical cell that drives electrical energy from spoutaneous redox reaction saking place within the cell.

1) Vol. of Cu to be plated $$=\cfrac{100\times 1}{10}=10cc$$

 wt. of cu required $$= 8.94\times 10=89.4$$

Moles of $$Cu = \cfrac{89.4}{63.5}$$

$$Cu^{2+}+2e^- \longrightarrow  Cu$$

Moles of electron $$= 89.4\times 10\times 2$$

amount. of electricity $$= \cfrac{89.4\times 10\times 2}{96500}$$ Farads

Or

2) Volume $$(V)$$ of $$Cu$$ $$= Area \times  thickness = \cfrac{10\times 10 \times  1}{102} = 1$$ .

$$d = \cfrac{m}{V}$$

$$m = d \times  V= 8.94 \times  1 = 8.94$$ .

Hence, Actual mass of $$Cu$$ to be plated is $$8.94 g$$ .

3) Now, $$Cu^{2+} + 2e^- \longrightarrow  Cu$$

So, $$2F \longrightarrow  63.5 g Cu$$

                    $$2 \times  96500 C \longrightarrow  63.5g$$ of $$Cu$$

                   $$P  C \longrightarrow  8.94 g Cu$$

                   $$P = 2 \times 96500 \times  \cfrac{8.94}{96500} = 27172.$$

Gold and platinum.
Corrosion of iron is a serious problem. Every year enormous amount of money is spent to replace damaged iron and steel structures.

Gold and platinum.
A thin impervious layer aluminium oxide froms a protective layer which protects the aluminium metals underneath from further damage.

$$Zn$$ occurs readily as ion whereas $$Cu$$ occurs more readily as metal in nature.

We should paint iron bars of the window,to protect them from rusting.Paint not only repairs rust spots, it can turn security bars into a decorative feature instead of making your home feel like a prison.

a. Tamarind is acidic. So aluminium corrodes by reacting with acid.
b. To reduce the contact with air.
c. Citrus fruit is acidic. It reacts with iron.

Yes, it prevents rusting. Iron nails, when kept inside kerosene, are protected from rusting. They are not in contact with atmospheric air or moisture.

At 298 K, $$E=0.675 \: V$$
At 308 K, $$E=0.675-6.5 \times 10^{-4} \times 10 =0.6685 \: V$$
$$\Delta G =-nFE$$
At 298 K, $$\Delta G=2 \times 96500 \times 0.675 = -130275 \: J/mol$$
At 308 K, $$\Delta G=2 \times 96500 \times 0.6685 = -129020.5 \: J/mol$$
$$\Delta G = \Delta H - T \Delta S$$
$$-130275 = \Delta H - 298 \Delta S$$ .......(1)
$$-129020.5 = \Delta H - 308 \Delta S$$ ......(2)
Subtract equation (2) from equation (1)
$$-1254.5=10 \Delta S$$
$$\Delta S = -125.45$$
$$\Delta H = -130275+298 (-125.45)=-167.65 \: kJ/mol =x$$
Hence, $$-100x=16765$$

For given cell reaction,

If $$E^o_1, E^o_2$$ and $$E^o_3$$ are standard oxidation potentials for $$Fe | Fe^{2+} , Fe^{2+}  | Fe^{3+}$$ and $$Fe| Fe^{3+}$$ , then $$E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}$$ . 

$$Cu^{2+}\, +\, e^{-}\, \rightarrow\, Cu^{\oplus}$$ ; 
$$E^{\ominus}\, =\, 1.77\, V$$
$$\Rightarrow\, \Delta G^{\ominus}_{1}\, =\, -\, 1\, \times\, F\, \times\, (1.77)\, =\, -1.77\, F$$
$$Cu^{2+}\, +\, e^{-}\, \rightarrow\, Cu$$
$$\Rightarrow\, \Delta G^{\ominus}_{2}\, =\, - 2\, \times\, F\, \times\, (- 0.34)\, =\, - 0.68 F$$
$$Cu^{oplus}\, +\, e^{-}\, \rightarrow\, Cu$$
$$\Delta G^{\ominus}\, =\, \Delta G^{\ominus}_{2}\, -\, \Delta^{\ominus}_{1}$$
$$\Rightarrow\, -1\, \times\, F\, \times\, E^{\ominus}\, =\, [0.68\, -\, (-1.77)]F\, \Rightarrow\, E^{\ominus}\, =\, - 2.45\, V$$

(2)
$$Al \rightarrow Al^{3+} + 3e^{-}$$

$$0.81\, =\, E^{\ominus}_{c}\, -\, E^{\ominus}_{a}\, -\, \displaystyle \frac{0.059}{2}{log}\, \frac{(0.02)}{(0.1)^{2}}$$ .

Let $$K_1$$ and $$K_2$$ represent the conductivities of two solutions A and B respectively.
Let $$x$$ be the cell constant.
For solution A,
The conductivity is the product of the conductance and the cell constant.
$$\displaystyle K_1 = \dfrac {x}{50}$$ ......(1)
For solution B
$$\displaystyle K_2 = \dfrac {x}{100}$$ ......(2)
Equal volumes of A and B are mixed. Both the solutions are doubly diluted.
Their individual contribution towards the conductivity of the mixture will be $$\displaystyle \dfrac {K_1}{2}$$ and $$\displaystyle \dfrac {K_2}{2}$$ respectively and the conductivity of the mixture will be $$\displaystyle \frac {K_1+K_2}{2}$$
For the mixture,
$$\displaystyle \dfrac {K_1+K_2}{2} =\dfrac {x}{R}$$ ......(3).
Here, $$R$$ is the resistance of the mixture.
Solving equations (1), (2) and (3), we get
$$\displaystyle R = 66.67 \approx 67$$ ohm.

$$\displaystyle E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^o_{Ag}-E^o_{Zn} = 0.344- (-0.76) = 1.104$$ V

According to Faraday's first law of electrolysis, the mass of the substance(W) deposited at an electrode is directly proportional to the amount of electricity or charge(Q) passed through.

A potential difference developing between the electrode and the electrolyte is known as electrode potential.

i. At anode : $$Ni \rightarrow Ni^{2+} + 2e^-$$  

At cathode :  $$[Ag^+ + e^- \rightarrow Ag] \times 2$$  

Cell reaction : $$Ni + 2Ag^+ \rightarrow Ni^{2+} + 2Ag$$  

  $$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$  

        $$= E^{\circ}_{Ag^+ / Ag} - E^{\circ}_{Ni^{2+} / Ni} = 0.80 \,V - (-0.25\,V)$$  

        $$E^{\circ}_{cell} = 1.05\,V$$  

ii. $$E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{n} \, log \dfrac{[Ni^{2+}]}{[Ag^+]^2}$$  

   Here , $$n = 2 , E^{\circ}_{cell} = 1.05 \,V , [Ni^{2+}] = 0.1\,M , [Ag^+] = 1.0\,M$$  

  $$E_{cell} = 1.05\,V - \dfrac{0.059}{2} \, log \dfrac{(0.1)}{(1)^2}$$  

  $$E_{cell} = 1.05 \,V - 0.0295 \, log 10^{-1}$$  

      $$= 1.05 + 0.0295\,V = 1.0795\,V$$