Class 12 Medical Chemistry - Electrochemistry

Lead is able to displace silver from AgNO $_3$ solution because its standard oxidation potential is higher than that of silver.
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Lead is a more active metal than silver and so, it can replace silver in a single replacement reaction. Standard oxidation potential of lead is higher than that of silver and so,

$\Delta G^o$ for the overall reaction is negative. Therefore, the reaction is feasible.

$Pb (s) + 2AgNO_3 (aq) \rightarrow Pb(NO_3)_2 (aq) + 2Ag (s)$

Calculate $E^{\ominus}_{cell}$ :
Given : $E^{\ominus}_{Ag^{\oplus}\, |\, Ag}\, =\, 0.80$ .

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

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

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

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

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

The standard electrode potential $(E^{\circ})$ for Daniel cell is $+1.1\ V$ . Calculate $\Delta G^{\circ}$ for the reaction.
$Zn(s) + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu(s)$
$(1\ F = 96500\ C/mol)$

$\Delta G =-nFE$

$= -2\times 96500\times 2.2$
=

$-212300\ J$
=

$-212.3\ kJ$

Jamila started noticing some kind of reddish brown particles around a photo frame, which is an old one. Why it could have happened and what is it?

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) Explain the effect of temperature and catalyst on the rate of a reaction.
(b) State Faraday's first law of electrolysis.
A solution of $CuSO_4$ is Electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

(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

$\displaystyle = \dfrac {I(A) \times t(s)}{96500}= \dfrac {1.5 \times 10 \times 60}{96500}= 0.009326$ 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

$\displaystyle =0.004663 \: mol \times 63.5 \: g/mol = 0.296 \: g$

State Faraday's laws of electrolysis.

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.

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

Resistance of a conductivity cell filled with 0.15 mol $L^{-1}$ NaCl solution is 50 $\Omega$ . If resistance of the same cell when filled with 0.02 mol $L^{-1}$ NaCl solution is 500 $\Omega$ , calculate the conductivity of 0.02 mol $L^{-1}$ NaCl solution. (The conductivity of 0.15 mol $L^{-1}$ NaCl solution is 1.5 s/m).

$\displaystyle \kappa = \dfrac {1}{R}\dfrac {l}{a}$
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

$\displaystyle \kappa = \dfrac {1}{R}\dfrac {l}{a}$

$\displaystyle \kappa = \dfrac {1}{500 \Omega} \times 75 /m = 0.150 S/m$

Distinguish between electrolytic cell and galvanic cell.

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

State Faraday's first law of electrolysis. For the electrode reaction $Zn^{2+}+2e^- \rightarrow Zn_{(s)}$ , what quantity of electricity in coulombs required to deposit one mole of zinc?

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.

Define : Standard electrode potential.

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

$25 C$ .

Define corrosion. What is meant by rust? Write the chemical formula of rust.

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 $Fe_2O_3.xH_2O$ .

Use $E^{\circ}$ values to calculate $\triangle G^{\circ}$ for the reaction
$Fe^{2+} + Ag^{+}\rightarrow Fe^{3+} + Ag$
$E^{\circ}_{Ag^{+}/Ag} = 0.80\ volt$ and $E^{\circ}_{Pt/ Fe^{3+}, Fe^{2+}} = 0.77\ volt$ .

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

$-2895\ J$ .

Solutions of two electrolytes $A$ and $B$ each having concentration of 0.2 M have conductivities $2 \times {10^{ - 2}},4 \times {10^{ - 4}}{\text{S/cm}}$ respectively. Which will offer greater resistance to the file of current and why?

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

$\implies R=\rho \cfrac{l}{A}$

Less is the conductivity, more is the resistivity, more is the resistance offer.

$\therefore$ Electrolyte

$B$ having less conductivity offer higher resistance.

What are galvanic cells? Explain the construction and working of galvanic cell with an example?

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

It generally consists of two different metals immerseal in an electrolytic,or of individual half-cells with different metals and twice ions in solution concerted by a salt-bridge or seperated by porous membrane.

Anode:

$Zn(s)$

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

Cathod:

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

$Cu(s)$

If the density of copper is $8.94 g/cm^3$ , the quantity of elctricity required to plate an area $100CM^2$ of the thickness of $10^{-2}$ cm using $CuSO_4$ solution as electrolyte is:

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

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 $Cu$

$P C \longrightarrow 8.94 g Cu$

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

Explain the term corrosion with examples.

When a metal is attacked by substances around it such as moisture, acids, etc. to turn into a new substance, it is said to corrode and this process is called corrosion.

Ex: The black coating on silver and the green coating on copper result from corrosion.

For the cells
$2Ag+Pt^{2+}\rightarrow 2Ag^++Pt; E^0=0.4$ V
$2Ag+F_2\rightarrow 2Ag^++2F^-; E^0=2.07$ V.
If the potential for the reaction $Pt\rightarrow Pt^{2+}+2e^-$ is assigned as zero, determine the potential for the following electrodes.
$Ag \rightarrow Ag^+ + e^-$

1713199_1738273_ans_4d23e1abf7ba4c70bb9f258cb4f9e314.jpg

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54\overset{o}{A}$ ?

1911547_1740357_ans_39780f420dea4f2398d706267054d926.jpg

Resistance of a conductivity cell filled with $0.1 M KCl$ solution is 100 $\Omega$ . If the resistance of the same cell when filled with $0.02 M KCl$ solution is 520 $\Omega$ . Calculate the conductivity and molar conductivity of 0.02 M $KCl$ solution. [Given conductivity of 0.1 M KCl is 1.29 $sin^{-1}$ ]

For 0.01 M KCl:

Cell constant = conductivity

$\times$ resistance

$= 1.29 \times 100 =129 m^{-1}$

For 0.02 M KCl:

Cell constant = conductivity

$\times$ resistance

129=conductivity

$\times$ 520

$\therefore$ conductivity=0.248

$S \, m^{-1}$

$2.48 \times 10^{-3} \, S \, Cm^{-1}$

Also Molar conductivity =

$conductivity \times \dfrac {1000}{M_(mole/1)}$

$2.48 \times 10^{-3} \times \dfrac {1000}{M_(0.02)}=124 \, S\, cm^2\, mol^{-1}$

or molar conducitivity =

$Conductivity(S\, m^{-1}) \times \dfrac {1}{M_{(mole/m^3)}}$

$\dfrac {0.248 \times 1}{0.02 \times 10^3}=12.4\times 10^{-3}=124\times 10^{-4}\, S\, m^2\, mole^{-1}$

[

$M=0.02 \dfrac {mole}{litre}=0.02 \dfrac {mole}{dm^3}=0.02 \dfrac {mole}{10^{-3}m^{3}}=0.02\times 10^3 \dfrac {mole}{m^3}]$

$\lambda =\dfrac {K\times 1000}{N}= \dfrac {1.11 \times ^{10^{-2}}\times 1000}{0.1}$

$111 \, S\, cm^2\, eq^{-1}=111 \times 10^4 \, S \, cm^2 \, eq^{-1}$

$\lambda =\dfrac {K_{(S\, m^{-1})}}{N _{(eq /m^3)}}=\dfrac{1.11}{0.1 \times 10^3}$

$111 \times 10^4 \, S \, cm^2 \, eq^{-1}$

The resistance of a solution 'A' is 50 ohm and that of the solution 'B' is 150 ohm, both solution being taken in the same conductivity cell. If equal volumes of solution A and B are mixed, what will be the resistance of the mixture using the same cell? (Assume that there is no increase in the degree of dissociation of A and B on mixing)

Let

$K_1$ and

$K_2$ be the conductivity of the solution A and B respectively and the constant of the cell be x.

$\therefore$ For solution A: conductivity=Conductance

$\times$ Cell costant

$K_1=\dfrac {1}{50} \times x$ ....(2)

For solution B: conductivity

$K_2=\dfrac {1}{150} \times x$ ....(2)

When the equal volume of A and B are mixed. Both the solution is doubly diluted, hence their individual contribution towards the conductivity of the mixture will be

$\dfrac {K_1}{2}$ and

$\dfrac {K_2}{2}$ respectively and the conductivity of the mixture will be

$\dfrac {1}{2} (K_1 + K_2)$

For the mixture :

$\dfrac {1}{2} (K_1 + K_2)=\dfrac {1}{R}\times x$

From Eqs. (1), (2) and (3)

$R=75 ohm$

If you leave a piece of iron in the open area for a few days, it acquires a film of brownish substance, called rust.
a) Do you think rust is different from iron?
b) Can you change rust back into iron by some simple method?

a) Rust is brown dust of 'iron oxide'. So, rust is a totally different form of iron.

b) Unfortunately, the damage caused by rust cannot be reversed. Once the metal has flaked away, you can only stop any more rust from occurring or replace it.

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is a serious problem.

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.

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is an advantage.

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

Among $Zn$ and $Cu$ , which would occur more readily in nature as metal and which as ion?

$Zn$ occurs readily as ion whereas

$Cu$ occurs more readily as metal in nature.

Give equation to show the chemical reactions of zinc and lead where it displaces copper from its compound.

$Zn(s)+CuSO_4(aq)\longrightarrow ZnSO_4(aq)+Cu(s)$

Copper sulphate

$Pb(s)+CuCl_2(aq)\longrightarrow PbCl_2(aq)+Cu(s)$

Copper chloride.

Why should we paint the iron bars of window?

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.

Give reasons for the following statements.
$\bullet$ Tamarind is not stored in aluminium vessels.
$\bullet$ It is a common practice to apply oil on iron articles and tools.
$\bullet$ Stainless steel knives instead of iron knives are preferred for cutting citrus fruits.

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.

Is there any advantage, if iron nails are kept inside kerosene. Why?

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

The e.m.f. of the cell $Cd\begin{vmatrix}Cd Cl_2\\ 1M\end{vmatrix} \begin{vmatrix}AgCl(s)\\ \end{vmatrix}Ag$ is 0.675 volt at 25 $^o$ C. The temperature coefficient of the cell is $-6.5 \times 10^{-4} Volt \space K^{-1}$ . If heat of the reaction is $x$ kJ then $-100x$ is :

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,

$-100x=16765$

$E^o_{cell}$ for reaction,
$4Al(s) + 3O_2(s) + 6H_2O + 4OH^- \rightarrow 4 [Al(OH)_4^-]$ is 2.73 V.
If $\triangle{G}_f^o$ for $OH^-$ and $H_2O$ are - 157 kJ mol $^{-1}$ and - 237.2 kJ mol $^{-1}$ . If $\triangle{G}_f^o$ for $[Al(OH)_4]^-$ is $x$ kJ.
$-x$ is:

For given cell reaction,

$\Delta G^o = - n E^o F$

$\therefore \Delta G^o = - 12 \times 2.73 \times 96500 J$

$= - 3.1613 \times 10^3 kJ$

$\begin{bmatrix}n = 12 \therefore 4 Al^o \rightarrow 4Al^{3+} + 12 e\rightleftharpoons \\ 3O_2^o + 12 e \rightarrow 6O^{2-}\end{bmatrix}$

Now for given reactions,

$\Delta G^o = 4 \times G_f^o [Al (OH)_4]^- -6 \times G_f^o [H_2O] - 4 \times G_f^o [OH^-]$

(Also note that

$G_f^o$ for elements is zero)

$- 3.1613 \times 10^3 = 4 \times G_f^o [Al (OH)_4]^- - 6 \times (- 237.2) - 4 \times (-157)$

$\therefore G_f^o [Al (OH)_4]^- = - 1303 kJ$

$-x= 1303$

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 $E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}$ . The value of n is:

$E^o_1, E^o_2$ and

$E^o_3$ are standard oxidation potentials for

$Fe | Fe^{2+} , Fe^{2+} | Fe^{3+}$ and

$Fe| Fe^{3+}$ , then

$E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}$ .

The value of n is 3.

The cell reactions for

$E^o_1, E^o_2$ and

$E^o_3$ require 2, 1 and 3 electrons respectively.

Hence,

$3E^o_3 = \displaystyle E^o_2 + 2E^o_1$ .

$E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{3}$ .

If for the following half-cell reactions :
$Cu^{2+}\, +\, e^{-}\, \rightarrow\, Cu^{\oplus}$ ;
$E^{\oplus}\, =\, 1.77\, V$
$Cu^{2+}\, +\, 2e^{-}\, \rightarrow\, Cu$ ;
$E^{\ominus}\, =\, -\, 0.34$
The $E^{\ominus}$ of the half-cell reaction will be (in V)
$Cu^{\oplus}\, +\, e^{-}\, \rightarrow\, Cu$

$Cu^{2+}\, +\, e^{-}\, \rightarrow\, Cu^{\oplus}$ ;

$E^{\ominus}\, =\, 1.77\, V$

$\Rightarrow\, \Delta G^{\ominus}_{1}\, =\, -\, 1\, \times\, F\, \times\, (1.77)\, =\, -1.77\, F$

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

$\Rightarrow\, \Delta G^{\ominus}_{2}\, =\, - 2\, \times\, F\, \times\, (- 0.34)\, =\, - 0.68 F$

$Cu^{oplus}\, +\, e^{-}\, \rightarrow\, Cu$

$\Delta G^{\ominus}\, =\, \Delta G^{\ominus}_{2}\, -\, \Delta^{\ominus}_{1}$

$\Rightarrow\, -1\, \times\, F\, \times\, E^{\ominus}\, =\, [0.68\, -\, (-1.77)]F\, \Rightarrow\, E^{\ominus}\, =\, - 2.45\, V$

$\Delta G$ for the reaction, $\displaystyle \frac{4}{3} Al + O_2 \rightarrow \frac{2}{3} Al_2O_3,$ is -772 kJ mol $^{-1}$ of $O_2$ . The minimum E.M.F. in volts required to carry out electrolysis of $Al_2O_3$ is:

(2)

$Al \rightarrow Al^{3+} + 3e^{-}$

$\displaystyle \dfrac{4}{3}\, mol\, of Al = \dfrac{4}{3} \times\,3\ mol\, e^{-} = 4\ mol\ e^{-}$

$n = 4$

Now,

$\Delta G = -nFE$

$-772\, \times 1000\, J\, =\, -4\, \times\, 96500\, \times E$

$\therefore E = \displaystyle \dfrac{772\, \times 1000}{4 \times 96500} = 2.0 V$

If the EMF of a cell :

$M\, |\, M^{2+}\, (0.02\, M)\, ||\, H^{+}\, (0.1\, M)\, |\, H_{2}$ (1 atm)

at $25^{\circ}\, C$ is 0.81V. Find the standard reduction potential of $M$ .
(Write a value to the nearest integer. If an answer is negative then write its magnitude)

From Nearst's equation-

$E_{cell} = E^o_{cell} - \cfrac{0.0593}{n} log \left( \cfrac{p_{H_2} \times [M^{2+}]}{[H^+]^2}\right)$

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

$\Rightarrow\, E^{\ominus}_{a}\, =\, 0\, -\, 0.81\, -\, 0.0089\, =\, -\, 0.8189\, V$

$E^{\ominus}_{M^{2+}\, |\, M}\, =\, -\, 0.8189\, V \approx -1\ V$

So, the correct answer is

$|-1|=1$ .

Ans

$1.$

The resistance of a solution A is 50 ohms and that of solution B is 100 ohms, both solutions being taken in the same conductivity cell. If equal volumes of solution A and B are mixed, the resistance (in ohms) of the mixture using the same cell will be $13x+2$ . Then $x$ is:
Assume that there is no increase in the degree of dissociation of A and B on mixing.
(write the value to the nearest integer)

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.

$R=13x +2 = 67$

Hence,

$x$ =5

In the button cells widely used in watches and other devices the following reaction takes place:
$Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)$
Determine $\Delta_{r} G^{\theta}$ and $E^{\theta}$ for the reaction.

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

$\displaystyle \Delta G^0 = -nFE^0_{cell} = - 2 \times 96500 \times 1.104 = -213072 \: J = -213 \: kJ$

Answer the following, in short, using a maximum $30$ words.
Explain Faraday's first law of electrolysis.

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.

$W\propto Q$

A galvanic cell consists of a metallic. zinc plate immersed in 0.1 M ${\text{Zn}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ solution and metallic plate of lead in 0.02 M ${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ solution. Calculate the emf of the cell write the chemical equation for the electrode reactions and represent the cell. (Given ${\text{E}}{^\circ _{Z{n^{2 + }},Zn}} = - 0.76V,\,E{^\circ _{P{b^{2 + }},Pb}} = 0.13V$ )

At anode

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

At cathode

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

Cell representation

$Zn / Zn(NO_3)_2 (0.1M) || Pb(NO_3)_2 (0.02M)/ Pb$

$E = E^{o} - \dfrac{0.0591}{n} log \dfrac {Zn^{2+}}{ Pb^{2+}}$

$E = (0.76 + 0.13) - \dfrac{0.0591}{2} log \dfrac {0.1}{ 0.02}$

$E = 0.08693 v$

According to the equation $Cu^{2+}+2e^- \rightarrow Cu$ , how much moles of copper are deposited when $965C$ of electricity is passed through a solution of $Cu^{2+}$ ions? $(1F=96500\,C)$

1413399_1324107_ans_d747bdb2e0f74ff7a9052d34094eb7d8.jpg

Why is an external emf of more than $2.2\ V$ required for the extraction of $Cl_{2}$ from brine?

For the reaction:

$2Cl^-(aq) + 2H_2O(l)\rightarrow 2OH^-(aq) + H_2(g) + Cl_2(g)$ ,

the value of

$\Delta G^{\ominus}$ is 422 kJ.

Using the equation

$\Delta G^{\ominus}=-nFE^{\ominus}$ the value of

$E^{\ominus}$ comes out to be -2.2 V.

Therefore extraction of

$Cl_2$ from brine will require an external emf of greater than 2.2 V.

The electrical resistance of a column of 0.05 M $NaOH$ solution of diameter 1 cm and length 50 cm is $5.55 \times 10^3$ ohm.Calculate its conductance, resistivity, conductivity and molar conductivity.

Given

$[NaOH]=0.05 M$ , l=50 cm, diameter of column=1 cm

$R=5.5 \times 10^3 ohm$

$\therefore$ Area of circular column

$=\pi r^2=3.14\times(0.5)^2=0.785 cm^2$

$\therefore$ Cell constant =

$\dfrac {1}{a}= \dfrac {50}{0.785} cm^{-1}$

$\therefore$ Conductance =

$\dfrac {1}{R}= \dfrac {1}{5.55\times 10^{-3}} ohm^{-1}$

$=181.82\, ohm^{-1}$

Resistivity

$= R\times \dfrac {a}{1}= 5.55 \times 10^3 \times \dfrac {0.785}{50}$

$=87.135 \, ohm \, cm$

Conductivity =

$\dfrac {1}{Resistivity}-\dfrac {1}{87.135}$

$0.01148 ohm^{-1}\, cm^{-1}$

Molar conductivity =

$\dfrac {conductivity \times 1000}{M}=\dfrac {0.01148 \times 1000}{0.005}$

$229.6 \, ohm^{-1}\, cm^2\, mol^{-1}$

The surface of copper gets tarnished by the formation of copper oxide. $N_2$ gas was passed to prevent the oxide formation during heating of copper at $1250$ K. However, the $N_2$ gas contains 1 mole per cent of water vapour as an impurity. The water vapour oxidises copper as per the reaction given below:

$2Cu(s) + H_2O(g) \rightarrow Cu_2O(s) + H_2(g)$

$P_{H_2}$ is the minimum partial pressure of $H_2$ (1 bar) needed to prevent the oxidation at 1250 K. The value of In $(P_{H_2})$ is .......

[Given: Total pressure = 1 bar, $R$ (universal gas constant) = $8 J K^{-1} mol^{-1}, ln (10) = 2.3, CuO(s)$ and $Cu_2O(s)$ are mutually immiscible.

At 1250 K: $2Cu(s) + \dfrac{1}{2}O_2(g) \rightarrow Cu_2O(s), \Delta G^0 = -78,000 \ J mol^{-1}$

$H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow H_2O(g); \Delta G^0 = 1,78,000 \ J mol^{-1}$

1737688_1738010_ans_76c2c7d014c94d6cafe19ae716de70b4.jpg

What is electrode potential?

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

A strip of nickel metal is placed in a $1$ molar solution of $Ni(NO_3)_2$ and a strip of silver metal is placed in a $1$ molar solution of $AgNO_3$ . An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltmeter .
(i) Write the balanced equation for the overall reaction occurring in the cell and calculate the cell potential .
(ii) Calculate the cell potential , E at $25^{\circ} \,C$ for the cell if the initial concentration of $Ni(NO_3)_2$ is $0.100$ molar and the initial concentration of $AgNO_3$ is $1.00$ molar .
$[E^{\circ}_{Ni^{2+} / Ni} = - 0.25\,V ; E^{\circ}_{Ag/Ag} = 0.80 \,V , log \, 10^{-1} = -1]$

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 ,

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

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

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

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

Electrochemistry - Class 12 Medical Chemistry - Extra Questions

Lead is able to displace silver from AgNO $_3$ solution because its standard oxidation potential is higher than that of silver.
If true enter 1, else enter 0

Calculate $E^{\ominus}_{cell}$ :
Given : $E^{\ominus}_{Ag^{\oplus}\, |\, Ag}\, =\, 0.80$ .

The standard electrode potential $(E^{\circ})$ for Daniel cell is $+1.1\ V$ . Calculate $\Delta G^{\circ}$ for the reaction.
$Zn(s) + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu(s)$
$(1\ F = 96500\ C/mol)$

Jamila started noticing some kind of reddish brown particles around a photo frame, which is an old one. Why it could have happened and what is it?

(a) Explain the effect of temperature and catalyst on the rate of a reaction.
(b) State Faraday's first law of electrolysis.
A solution of $CuSO_4$ is Electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

State Faraday's laws of electrolysis.

Distinguish between electrolytic cell and galvanic cell.

State Faraday's first law of electrolysis. For the electrode reaction $Zn^{2+}+2e^- \rightarrow Zn_{(s)}$ , what quantity of electricity in coulombs required to deposit one mole of zinc?

Define : Standard electrode potential.

Define corrosion. What is meant by rust? Write the chemical formula of rust.

Use $E^{\circ}$ values to calculate $\triangle G^{\circ}$ for the reaction
$Fe^{2+} + Ag^{+}\rightarrow Fe^{3+} + Ag$
$E^{\circ}_{Ag^{+}/Ag} = 0.80\ volt$ and $E^{\circ}_{Pt/ Fe^{3+}, Fe^{2+}} = 0.77\ volt$ .

Solutions of two electrolytes $A$ and $B$ each having concentration of 0.2 M have conductivities $2 \times {10^{ - 2}},4 \times {10^{ - 4}}{\text{S/cm}}$ respectively. Which will offer greater resistance to the file of current and why?

What are galvanic cells? Explain the construction and working of galvanic cell with an example?

If the density of copper is $8.94 g/cm^3$ , the quantity of elctricity required to plate an area $100CM^2$ of the thickness of $10^{-2}$ cm using $CuSO_4$ solution as electrolyte is:

Explain the term corrosion with examples.

For the cells
$2Ag+Pt^{2+}\rightarrow 2Ag^++Pt; E^0=0.4$ V
$2Ag+F_2\rightarrow 2Ag^++2F^-; E^0=2.07$ V.
If the potential for the reaction $Pt\rightarrow Pt^{2+}+2e^-$ is assigned as zero, determine the potential for the following electrodes.
$Ag \rightarrow Ag^+ + e^-$

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54\overset{o}{A}$ ?

If you leave a piece of iron in the open area for a few days, it acquires a film of brownish substance, called rust.
a) Do you think rust is different from iron?
b) Can you change rust back into iron by some simple method?

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is a serious problem.

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is an advantage.

Among $Zn$ and $Cu$ , which would occur more readily in nature as metal and which as ion?

Give equation to show the chemical reactions of zinc and lead where it displaces copper from its compound.

Why should we paint the iron bars of window?

Give reasons for the following statements.
$\bullet$ Tamarind is not stored in aluminium vessels.
$\bullet$ It is a common practice to apply oil on iron articles and tools.
$\bullet$ Stainless steel knives instead of iron knives are preferred for cutting citrus fruits.

Is there any advantage, if iron nails are kept inside kerosene. Why?

The e.m.f. of the cell $Cd\begin{vmatrix}Cd Cl_2\\ 1M\end{vmatrix} \begin{vmatrix}AgCl(s)\\ \end{vmatrix}Ag$ is 0.675 volt at 25 $^o$ C. The temperature coefficient of the cell is $-6.5 \times 10^{-4} Volt \space K^{-1}$ . If heat of the reaction is $x$ kJ then $-100x$ is :

$E^o_{cell}$ for reaction,
$4Al(s) + 3O_2(s) + 6H_2O + 4OH^- \rightarrow 4 [Al(OH)_4^-]$ is 2.73 V.
If $\triangle{G}_f^o$ for $OH^-$ and $H_2O$ are - 157 kJ mol $^{-1}$ and - 237.2 kJ mol $^{-1}$ . If $\triangle{G}_f^o$ for $[Al(OH)_4]^-$ is $x$ kJ.
$-x$ is:

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 $E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}$ . The value of n is:

If for the following half-cell reactions :
$Cu^{2+}\, +\, e^{-}\, \rightarrow\, Cu^{\oplus}$ ;
$E^{\oplus}\, =\, 1.77\, V$
$Cu^{2+}\, +\, 2e^{-}\, \rightarrow\, Cu$ ;
$E^{\ominus}\, =\, -\, 0.34$
The $E^{\ominus}$ of the half-cell reaction will be (in V)
$Cu^{\oplus}\, +\, e^{-}\, \rightarrow\, Cu$

$\Delta G$ for the reaction, $\displaystyle \frac{4}{3} Al + O_2 \rightarrow \frac{2}{3} Al_2O_3,$ is -772 kJ mol $^{-1}$ of $O_2$ . The minimum E.M.F. in volts required to carry out electrolysis of $Al_2O_3$ is:

If the EMF of a cell :

$M\, |\, M^{2+}\, (0.02\, M)\, ||\, H^{+}\, (0.1\, M)\, |\, H_{2}$ (1 atm)

at $25^{\circ}\, C$ is 0.81V. Find the standard reduction potential of $M$ .
(Write a value to the nearest integer. If an answer is negative then write its magnitude)

In the button cells widely used in watches and other devices the following reaction takes place:
$Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)$
Determine $\Delta_{r} G^{\theta}$ and $E^{\theta}$ for the reaction.

Answer the following, in short, using a maximum $30$ words.
Explain Faraday's first law of electrolysis.

According to the equation $Cu^{2+}+2e^- \rightarrow Cu$ , how much moles of copper are deposited when $965C$ of electricity is passed through a solution of $Cu^{2+}$ ions? $(1F=96500\,C)$

Why is an external emf of more than $2.2\ V$ required for the extraction of $Cl_{2}$ from brine?

The electrical resistance of a column of 0.05 M $NaOH$ solution of diameter 1 cm and length 50 cm is $5.55 \times 10^3$ ohm.Calculate its conductance, resistivity, conductivity and molar conductivity.

What is electrode potential?

Class 12 Medical Chemistry Extra Questions

Electrochemistry - Class 12 Medical Chemistry - Extra Questions

Lead is able to displace silver from AgNO3_3 solution because its standard oxidation potential is higher than that of silver. If true enter 1, else enter 0

Calculate E⊖cell E^{\ominus}_{cell} :Given : E⊖Ag⊕|Ag=0.80E^{\ominus}_{Ag^{\oplus}\, |\, Ag}\, =\, 0.80.

The standard electrode potential (E∘)(E^{\circ}) for Daniel cell is +1.1 V+1.1\ V. Calculate ΔG∘\Delta G^{\circ} for the reaction.Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu(s)(1 F=96500 C/mol)(1\ F = 96500\ C/mol)

Jamila started noticing some kind of reddish brown particles around a photo frame, which is an old one. Why it could have happened and what is it?

(a) Explain the effect of temperature and catalyst on the rate of a reaction.(b) State Faraday's first law of electrolysis.A solution of CuSO4CuSO_4 is Electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

State Faraday's laws of electrolysis.

Distinguish between electrolytic cell and galvanic cell.

State Faraday's first law of electrolysis. For the electrode reaction Zn2++2e−→Zn(s)Zn^{2+}+2e^- \rightarrow Zn_{(s)}, what quantity of electricity in coulombs required to deposit one mole of zinc?

Define : Standard electrode potential.

Define corrosion. What is meant by rust? Write the chemical formula of rust.

Use E∘E^{\circ} values to calculate △G∘\triangle G^{\circ} for the reactionFe2++Ag+→Fe3++AgFe^{2+} + Ag^{+}\rightarrow Fe^{3+} + AgE∘Ag+/Ag=0.80 voltE^{\circ}_{Ag^{+}/Ag} = 0.80\ volt and E∘Pt/Fe3+,Fe2+=0.77 voltE^{\circ}_{Pt/ Fe^{3+}, Fe^{2+}} = 0.77\ volt.

Solutions of two electrolytes AA and BB each having concentration of 0.2 M have conductivities 2×10−2,4×10−4S/cm2 \times {10^{ - 2}},4 \times {10^{ - 4}}{\text{S/cm}} respectively. Which will offer greater resistance to the file of current and why?

What are galvanic cells? Explain the construction and working of galvanic cell with an example?

If the density of copper is 8.94g/cm38.94 g/cm^3, the quantity of elctricity required to plate an area 100CM2100CM^2 of the thickness of 10−210^{-2} cm using CuSO4CuSO_4 solution as electrolyte is:

Explain the term corrosion with examples.

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54oA1.54\overset{o}{A}?

If you leave a piece of iron in the open area for a few days, it acquires a film of brownish substance, called rust.a) Do you think rust is different from iron?b) Can you change rust back into iron by some simple method?

Which two metals do not corrode easily ? Give example in each case to support thatCorrosion of some metals is a serious problem.

Which two metals do not corrode easily ? Give example in each case to support that Corrosion of some metals is an advantage.

Among ZnZn and CuCu, which would occur more readily in nature as metal and which as ion?

Give equation to show the chemical reactions of zinc and lead where it displaces copper from its compound.

Why should we paint the iron bars of window?

Give reasons for the following statements.∙\bullet Tamarind is not stored in aluminium vessels.∙\bullet It is a common practice to apply oil on iron articles and tools.∙\bullet Stainless steel knives instead of iron knives are preferred for cutting citrus fruits.

Is there any advantage, if iron nails are kept inside kerosene. Why?

If Eo1,Eo2E^o_1, E^o_2 and Eo3E^o_3 are standard oxidation potentials for Fe|Fe2+,Fe2+|Fe3+Fe | Fe^{2+} , Fe^{2+} | Fe^{3+} and Fe|Fe3+Fe| Fe^{3+}, then Eo3=Eo2+2Eo1nE^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}. The value of n is:

ΔG\Delta G for the reaction, 43Al+O2→23Al2O3,\displaystyle \frac{4}{3} Al + O_2 \rightarrow \frac{2}{3} Al_2O_3, is -772 kJ mol−1^{-1} of O2O_2. The minimum E.M.F. in volts required to carry out electrolysis of Al2O3Al_2O_3 is:

If the EMF of a cell :M|M2+(0.02M)||H+(0.1M)|H2M\, |\, M^{2+}\, (0.02\, M)\, ||\, H^{+}\, (0.1\, M)\, |\, H_{2} (1 atm) at 25∘C25^{\circ}\, C is 0.81V. Find the standard reduction potential of MM.(Write a value to the nearest integer. If an answer is negative then write its magnitude)

In the button cells widely used in watches and other devices the following reaction takes place:Zn(s)+Ag2O(s)+H2O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)Determine ΔrGθ\Delta_{r} G^{\theta} and EθE^{\theta} for the reaction.

Answer the following, in short, using a maximum 3030 words.Explain Faraday's first law of electrolysis.

According to the equation Cu2++2e−→CuCu^{2+}+2e^- \rightarrow Cu, how much moles of copper are deposited when 965C965C of electricity is passed through a solution of Cu2+Cu^{2+} ions? (1F=96500C)(1F=96500\,C)

Why is an external emf of more than 2.2 V2.2\ V required for the extraction of Cl2Cl_{2} from brine?

The electrical resistance of a column of 0.05 M NaOHNaOH solution of diameter 1 cm and length 50 cm is 5.55×1035.55 \times 10^3 ohm.Calculate its conductance, resistivity, conductivity and molar conductivity.

What is electrode potential?

Class 12 Medical Chemistry Extra Questions

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Lead is able to displace silver from AgNO $_3$ solution because its standard oxidation potential is higher than that of silver.
If true enter 1, else enter 0

Calculate $E^{\ominus}_{cell}$ :
Given : $E^{\ominus}_{Ag^{\oplus}\, |\, Ag}\, =\, 0.80$ .

The standard electrode potential $(E^{\circ})$ for Daniel cell is $+1.1\ V$ . Calculate $\Delta G^{\circ}$ for the reaction.
$Zn(s) + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu(s)$
$(1\ F = 96500\ C/mol)$

(a) Explain the effect of temperature and catalyst on the rate of a reaction.
(b) State Faraday's first law of electrolysis.
A solution of $CuSO_4$ is Electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

State Faraday's first law of electrolysis. For the electrode reaction $Zn^{2+}+2e^- \rightarrow Zn_{(s)}$ , what quantity of electricity in coulombs required to deposit one mole of zinc?

Use $E^{\circ}$ values to calculate $\triangle G^{\circ}$ for the reaction
$Fe^{2+} + Ag^{+}\rightarrow Fe^{3+} + Ag$
$E^{\circ}_{Ag^{+}/Ag} = 0.80\ volt$ and $E^{\circ}_{Pt/ Fe^{3+}, Fe^{2+}} = 0.77\ volt$ .

Solutions of two electrolytes $A$ and $B$ each having concentration of 0.2 M have conductivities $2 \times {10^{ - 2}},4 \times {10^{ - 4}}{\text{S/cm}}$ respectively. Which will offer greater resistance to the file of current and why?

If the density of copper is $8.94 g/cm^3$ , the quantity of elctricity required to plate an area $100CM^2$ of the thickness of $10^{-2}$ cm using $CuSO_4$ solution as electrolyte is:

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54\overset{o}{A}$ ?

If you leave a piece of iron in the open area for a few days, it acquires a film of brownish substance, called rust.
a) Do you think rust is different from iron?
b) Can you change rust back into iron by some simple method?

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is a serious problem.

Which two metals do not corrode easily ? Give example in each case to support that
Corrosion of some metals is an advantage.

Among $Zn$ and $Cu$ , which would occur more readily in nature as metal and which as ion?

Give reasons for the following statements.
$\bullet$ Tamarind is not stored in aluminium vessels.
$\bullet$ It is a common practice to apply oil on iron articles and tools.
$\bullet$ Stainless steel knives instead of iron knives are preferred for cutting citrus fruits.

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 $E^o_3 = \displaystyle \frac{E^o_2 + 2E^o_1}{n}$ . The value of n is:

$\Delta G$ for the reaction, $\displaystyle \frac{4}{3} Al + O_2 \rightarrow \frac{2}{3} Al_2O_3,$ is -772 kJ mol $^{-1}$ of $O_2$ . The minimum E.M.F. in volts required to carry out electrolysis of $Al_2O_3$ is:

If the EMF of a cell :

$M\, |\, M^{2+}\, (0.02\, M)\, ||\, H^{+}\, (0.1\, M)\, |\, H_{2}$ (1 atm)

at $25^{\circ}\, C$ is 0.81V. Find the standard reduction potential of $M$ .
(Write a value to the nearest integer. If an answer is negative then write its magnitude)

In the button cells widely used in watches and other devices the following reaction takes place:
$Zn(s) + Ag_{2}O(s) + H_{2}O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^{-}(aq)$
Determine $\Delta_{r} G^{\theta}$ and $E^{\theta}$ for the reaction.

Answer the following, in short, using a maximum $30$ words.
Explain Faraday's first law of electrolysis.

According to the equation $Cu^{2+}+2e^- \rightarrow Cu$ , how much moles of copper are deposited when $965C$ of electricity is passed through a solution of $Cu^{2+}$ ions? $(1F=96500\,C)$

Why is an external emf of more than $2.2\ V$ required for the extraction of $Cl_{2}$ from brine?

The electrical resistance of a column of 0.05 M $NaOH$ solution of diameter 1 cm and length 50 cm is $5.55 \times 10^3$ ohm.Calculate its conductance, resistivity, conductivity and molar conductivity.