In the last lecture we have solved some problems,

those problems we have used the triangle where you have the relation between del G 0, equilibrium

constant and reduction potential. And also, we have seen that if you can stack all the

standard reduction potential in the form of a series with respect to hydrogen electrode,

then you also get a standard reduction, standard reduction potential series. And in that if

you have 0 potential for the standard hydrogen electrode, then the top of that electrode

would be top of that hydrogen electrode. Let us say, if you have 0 as this H plus plus

E half H 2, this is my 0 volt and if you stack all the E 0 and this is rho in the form of

reduction potential, then you get a series and that series you can judge. If the metals

or the reduction potential above 0 voltage, those are active, those are noble and 0, 0

voltage, all the metal potential, all the reduction potential of other reduction reactions

will be considered as active. Now, this standard reduction potential says

one is whether metal is active or noble, that means if the metal potential is above 0 voltage

or below 0 voltage, if this E 0 should be greater than 0 volt and here it is E 0 should

be less than 0 volt. Now, the second information what we get from

standard reduction potential series, if their metal corrosion can be understood readily.

For example, if you have copper and zinc and if you have copper plus plus ion in contact

with zinc, then zinc will come out plus plus and copper will deposit. So, this would be

the reduction, this would be oxidation and this is also through because we know, that

for this E 0 Zn plus plus Zn 2 Z n, that means, reduction potential is minus 0.761 volt. And

in this case, E 0 Cu plus plus Cu equal to plus 0.34 volt. So, that means, this is more

than this, hence copper ion will deposit zinc ion; zinc will corrode in the form of zinc

ion. So, this information we can get for pure metal. Now, another case is, it provides quick

glimpse of which material will be more prone to corrosion. That means, in this case zinc

has the more proneness to corrosion and in this case copper has less proneness to corrosion,

because zinc has higher reduction potential or higher reduction ability.

Now, there are few disadvantages and most mostly, basically, it gives you something,

some theoretical basis, that which one is noble, which one is active because it based

on thermodynamic data. Because del G, why this del E, what you are getting del E or

E 0? What you are measuring with respect to standard hydrogen electrode in case of pure

metal because here, the activity of the ionic species, all the activities will be one. So,

this is basically nothing but del G 0 by n F minus, because this relation, from this

relation you are getting E 0, that means, del G 0 is purely thermodynamic data.

So, it is thermodynamics, nothing about kinetics. So, you will not get any information about

the rate of this process. You know, that this is possible, but you cannot tell from this

relation or form this data at what way zinc will corrode and what way copper will deposit,

this information we cannot get. Now, coming to another point, let us say,

if you are considering the corrosion of an alloy, let us say copper-zinc alloy, so in

that case or brass, in that case this information will not tell you what would be the situation

here, whether copper will at all corrode or zinc will corrode, those information we cannot

get from this table. Now, another point is, this is a standard value, so the temperature

is 298 Kelvin. So, if you consider different temperature, then also you cannot get any

information. Let us say I am giving one information, one example, zinc and iron, if you combine

this, iron has E 0 Fe plus plus Fe, its value is minus 0.44 volt. Now, if you combine this

zinc and iron in HCl solution, then from this is pretty clear, that this has higher reduction

potential than zinc, so iron should corrode more than zinc. That actually, it would be

reversed if you club, if you galvanically couple iron and zinc in HCl solution. So,

in that case, zinc corrosion would be less, iron corrosion would be more. But still, from

this information, from this data, at least we cannot, we can, cannot say, that the reverse

is happening. So, that information we cannot get. So, these are, that means, the practical

corrosion information we cannot get out of this standard reduction potential series.

But it has huge importance, that importance of where, the importance of this has been

find out in the problem where we have tried to see nickel, whether nickel would corrode

in a deaerated H2O medium where if you consider the last class, last lecture problem where

we have seen, that whether nickel, we have to find out theoretically, whether nickel

can corrode as nickel hydroxide in deaerated H2O and with few information.

For example, we know k SP of Ni OH whole 2, we know that value, we know what are the reactions,

rate of reactions. So, nickel, if nickel has to corrode, so it should go this way and another

cathodic reaction would happen like this and we know the pH of the solution. These are

the problem, this was the problem and we had to find out whether actually, in this condition,

set of conditions nickel would corrode or not. And we have found out, that nickel does

not want to corrode in this condition, in these set of conditions because we have found

out, that del G for this process would be positive. If it is positive, that means, it

is non-spontaneous, that means, spontaneously this reaction would not happen. So, rather

we can say that nickel cannot corrode in this set of conditions.

Now, nickel hydroxide, now you see what are the products, what are the, what are the reactants,

what are the products or rather what are those species in this condition. This is one species,

nickel. This is another species, because nickel hydroxide is forming, initially it forms nickel,

nickel plus plus ion and then nickel plus pus ion will react with H2O and forms Ni OH

whole 2, then also we have H plus, we have H 2. So, these are the species. So, that means,

if we, if we put all the species one by one, one is nickel, second is nickel plus plus,

third is nickel hydroxide, fourth is H 2 O, fifth is H plus. So, these are the species

in deaerated H 2 O. Now, let us say I have aerated medium. Now,

if I have aerated medium, then I have to add another species, which is nothing but dissolve

oxygen, O 2. So, these are the species those are available in the medium, if it is, this

is in case of aerated, aerated solution. So, from this, in the problem we have just studied

what would be the stability of different phases as a function of p H or potential? So, that

means, we can, because when we try to find out del G we had to find out what is the potential

difference between in the cell or between the reaction of, reduction, reduction reaction

and oxidation reaction. So, that is one information we are getting and with the help of stability,

with the help of pH, with the help of solubility product of the precipitate or the phase that

is forming on the nickel surface. So, now if we try to see different reactions

that can be possible combining these six species, let me note down all those reactions. One

reaction is Ni plus plus plus 2e going to Ni, this reduction process can take place

or in other way round we can say, that nickel is also going back and forming Ni plus, that

oxidation process can also take place. Now, but I preferred to write it in terms of reduction

because that would help me to find out what would be my reduction potential.

Now, second reaction that is possible, Ni plus H 2 O, Ni plus H 2 O going to NiO plus

2H plus plus 2e or I would prefer to write it in other way around, let me write it in

this way because I have to write it in terms of reduction, so I can write it as Ni plus

H 2 O, this is another reaction. Third reaction that are possible, Ni plus plus plus H 2 O

going to NiO 2 H 2 O going to Ni OH whole 2 plus 2 H plus, this is also possible; I

am just writing all the possibilities . Now, fourth is again this Ni O, which is the

product of the reaction between nickel and H 2 O. In that case I can also form NiO, can

react with H 2 O and from Ni OH whole 2. So, this is also possible. So, these four reactions,

that are possible. What I can find, there are other reactions, but let me put all the

simple reactions, these are the simple reactions. Now, what are the other reactions? Five, H

plus plus e half H 2, this is possible. Now, 6th, now this is in case of deaerated media.

Now, if it is aerated, now you have another species, O 2, O 2 can react with 4 H plus

plus 4e, it can go to 2 H 2 O. 7th, another reaction that is possible, H 2 O plus 4e going

to 4 OH minus. Now, one more reaction, that I can think of, 2 H 2 O plus 2e going to H

2 plus 2 H minus. These are eight reactions, simple reactions that I can think of considering

all those six species. Now, let me see what information, what thermodynamic

information I can get from this reactions. Now, one thermodynamic reaction is, I can

find out what would be the mu of all those species, mu of all those species; that means

I can get mu of all species. Now, let me note down all those mu’s because those are thermodynamic

data. Now, let me note down all those mu’s first.

First is, now when we try to find out mu, let me put it in such a manner, that I am,

I am having some data for mu 0 or the standard chemical potential for all the species, all

the individual species. Now, this I can find, I can have some data, I can have data mu 0,

I can have data. Now, then what else I can have? Data for mu 0 H plus, mu 0 H 2, mu 0

O 2, mu 0 H 2 O, mu 0 OH minus. So, these many mu’s I can find it.

Now, convention is when you consider a pure metal, this is considered to be 0; again,

this is 0; again, this is considered to be 0 again because is a pure gas. Now, I can

have this one, this is also assumed to be 0 joules per mole. Now, I can have other data,

let me have other data, Ni plus plus is minus 46398 joule per mole. This is NiO H 2O is

minus 23694.2 joule per mole. Then I have data for Ni OH whole 2 minus, 108.300; this

is minus 47452694 joule per mole. Then, I

can have a data for, data for NiO, NiO would

be, just let me find out, you just find out, this is 51610 into 4.18, this is in calorie,

calorie per mole. So, it would be equal to minus, it would be equal to, let me, I will

just calculate, if you calculate it, let me see the, now this value is 51610 into 4.18

equal to minus 215729.8 joule per mole. So, let me remove this part. So, this is the value

for this. Now, I have the value, I, the value that is remain is mu 0 OH minus, mu 0 OH minus

data, let me also put that data. So, the, I, one data is missing, that is for mu 0 OH

minus, that is minus 157147.1 joule per mole. Now, with these datasets let me find out how

many, what are the data I can make out of these standard data? From this, this is you see, that this is an

electrochemical reaction, a reduction reaction. So, I can find out what is e of, for this

reaction, e for this reaction, E 0 rather because I have considered all those standard

state and it is considered at 25 degree Celsius per unit pressure on atmosphere pressure.

Now, for this is also an electrochemical reaction, at the same time you see the difference between

this and this, you have considered here H plus ion. Now, at the same time you have two

electrons, so that means, it is electron accepting reaction or reduction reaction. At the same

time, this reaction will depend on the concentration of H plus ion in the solution or in the aqueous

medium and this concentration of H plus ion in the aqueous medium or H 2 O can also be

related to pH of the solution. That means, this relation depends on potential, as well

as, pH of the solution. Now, this reaction, you say here you do not have an electron associated

with it. So, rather it is a reaction, which is not an electrochemical reaction.

So, now here it is basically involving, this reaction is involving Ni plus plus with H

2 O and forming Ni OH whole 2 and H plus ion. So, now, again this reaction is pH dependent,

but of course, this would be definitely potential independent. Now, again coming to this, if

you see these two, between, difference between three and four, three is pH dependent, potential

independent, but this is potential independent, as well as, pH independent. So, pH of the

solution, as well as potential does not matter to have some data out of it. Now, this is

also electron involved, as well as H plus. So, this is pH, as well as potential dependent.

This is also, say H plus and electron, so this is also potential and pH dependent.

This is another reaction where you have OH minus ion. So, OH minus ion you can also reflect

in terms of basicity, or in other way, you can also put it as in terms of pH. So, let

us say pH 7 that means neutral solution. So, in that case, pH, you can have pH, at the

same time you can have pOH. So, pOH, if it is 7, so this is 7, pH 7, so in that case

it would be also 7. So, neutral solution, pH, pOH is also 7, pOH is 7.

So, if you know this, you can also know this from the solubility product of H 2 O. Now,

or the dissolution dissociation constant for this H 2 O, again this one, again it is coming

to be a potential, as well as, pH dependent. So, you have few reactions, one reaction is,

one set of reactions, now let me come to name all those reactions, let me put in other,

let me put in such a way, that we have three set of reactions, one is pH dependent only,

pH dependent, rather let me put it, only pH dependent. That means, let us say this reaction,

pH of the solution will decide whether nickel plus plus will go to form Ni OH whole 2 in

the presence of H 2 or not, this is only pH dependent. Now, second set of reactions, which

are potential dependent, which are potential dependent and third set of reactions where

you have pH, as well as, potential, potential dependent.

So, three sets, let us say, let me see, this one is definitely pH independent only potential

dependent. So, let me put the number, this is, it falls under category of two. Second,

same thing, let me put it, this reaction here, pH as well as potential. So, it falls under

category 3. Now, this falls under category 1 because it is only pH dependent. Now, this

one, now there is one set of reaction, which is not pH or not potential dependent. So,

we have one set, which is not pH or potential, potential dependent. So, this is falling under

category 4. Now, again I can also put them in different categories, this is pH, as well

as, potential. So, category number 3, this is pH plus potential category, number 3. This

is again also OH minus; I can put it in terms of pH.

So, this is also potential because it involves electron, so electron and pH, so this is also

falling under category number 3. This is also falling under category number 3 because it

also involves electron, as well as, OH minus ion and OH minus ion you can put it in terms

of H plus ion or pH. So, now you see, whatever reactions that we have thought of and those

are happening in H 2 O solution in the presence of H plus and O 2, dissolved oxygen, these

many reaction for the time we can think of, which make, which leads to nickel ion plus

nickel plus plus formation, NiO formation, Ni OH whole 2 formation or the other way around. So, if we think this is eight reactions can

we put it on a diagram, where you have two axis, one is potential, one is pH, because

all those reactions, except this reaction, number 4, I can put them on this E versus

pH diagram, because all the reactions either they are potential dependent or pH dependent

or pH or potential dependent. So, when I put them in this diagram, E versus

pH, then I can specify, that which section has Ni plus plus, which section I have Ni,

which section I have H 2 O, which section I have H plus, which section I have NiO, all

those species I can specify on this diagram. Now, when I specify in this diagram, that

time I construct a sort of, now when I put them these reactions on this diagram, then

I construct Pourbaix diagram. This is known as Pourbaix diagram. So, E versus p H diagram

is nothing but the Pourbaix diagram where it is a graphical representation of all those

reactions that are happening in H 2 O in the presence of oxygen or H plus ion or both.

So, it is basically a kind of stability of a metal and its corrosion products as a function

of potential and pH, acidity or alkalinity of the aqueous solution.

Now, stability of a metal, because we have already seen in case of nickel corrosion,

that nickel forms, forms any Ni OH whole 2 and if it forms, if it forms Ni OH whole 2,

then we can also judge from this available data, that is a solubility product pH of the

median, whether the further reaction from nickel to Ni plus plus would be happening

or not fine and that is based on thermodynamic data.

Now, similar way I can plot these reactions on this diagram and also I would see, that

for a particular region of potential and pH, some phase or some species will have higher

tendency to form because I am, I am using this tendency word because we are judging

from thermodynamics. It cannot tell you whether actually that thing would happen or not, but

it will definitely give you some idea or tendency that this phase would or this phase or species

would form or not. Now, let me plot these reactions on this diagram one by one. So, first let me consider this reaction, this

is pH and this is E or E rather E 0 since I would be using all those standard reduction

potential, standard chemical potential and if I use standard chemical potential, then

this axis, if it is potential dependent, would come as E 0. Now, if you consider this reaction,

this reaction can be written in terms of, if you consider what would be the potential,

I can write in the form of E equal to Ni plus plus Ni, that is a reduction potential, would

be E 0 minus plus RT electron 2F ln Ni plus plus. Now, R is 8.314 joule per mole per Kelvin,

T is 298 Kelvin because I have considered 25 degree Celsius, 298 Kelvin and F is 96500,

F is, this is T, this is R, which is universal gas constant and this is F is equal to 96500

coulomb per gram equivalent. Now, if I put all those values here, then I would get E

0 plus 0.059, roughly 0.059 log concentration of nickel plus plus. Now, if I take from,

if I convert this we would get this. Instead of ln I am putting l o g, log and this is

E Ni plus plus Ni. Now, if I, if I have the situation where nickel

plus plus ion concentration is maintained at 1, this is 1 concentration, if the concentration

is 1, then E, the potential for this reaction when the concentration of Ni plus plus ion

is 1 unit. So, then I will get E Ni plus plus equal to E 0, which is standard state, standard

reduction potential for nickel. Now, how would we get standard reduction potential for nickel?

Then again I can have mu 0 Ni mu 0 Ni plus plus.

Now, I know what is the value of mu 0 Ni plus plus and mu 0 Ni, now what would be my, from

this I can calculate what would be my change in free energy and since I am considering

del G mu 0, so I will be considering del G 0. So, it would be mu 0, this would be mu

0, mu 0, Ni minus mu 0 Ni plus plus because this is my product, this is my reactant. So,

if this is my product, this is my reactant, so let me put it in product, this is reactant.

So, mu 0 nickel, which is product mu 0 Ni plus plus, which is reactant.

Now, this would, you will get, let me put the value plus 46398 joule per mole. Now,

if I have this what would be my potential? Potential, I can put it in this fashion, del

G 0 equal to minus nFE 0, n is 2 here, so E 0 would be del G 0 by 2 into 96500 and there

would be a minus sign. And it would be, if you find out this value, then it would come

as 46398 divided by 2 divided by 96500, it is coming, it as coming as 0.24 minus volt.

So, E 0 Ni plus plus Ni, the potential is coming as minus 0.24 volt, whatever data we

have. So, now you see that you have found out E 0 when the nickel ion concentration

is 1. Now, if you know the E 0 with the thermodynamic

data available here, then you have a relation, which is nothing but E Ni plus plus Ni equal

to E 0 Ni plus plus Ni. If this one is 1 and if it is not 1, then this is my relation,

this is my relation and we have also found out value of E 0 Ni plus plus Ni with the

thermodynamic data what we have now on this board.

Now, this is an equation, which relates potential to another standard potential and this is,

there is no pH term that is coming into this equation and if we would like to plot this

equation on this diagram, this should definitely be a horizontal line with respect to pH axis

because it is a pH independent reaction. Now, let me plot this on this diagram, let me put

that diagram there, so you can understand better because whatever calculation I will

be doing, that would be done on this side and the final plot would be here. So, this is my pH, this is my potential and

this potential is nothing but the reduction potential, here I can write E, Ox, Red. Now,

if I know this value minus 0.24, let me, this is 0.24 and this is the axis, which is input.

Now, if I would like to plot this equation on this diagram, then it should be a line,

which is parallel to this axis and if, we would like to see the nickel ion concentration.

If we change this nickel ion concentration to, from 1 to 10 to the power minus 2, if

I put in 10 to the power minus 2 here, so then E of this Ni plus plus, Ni of that, for

that reaction if the concentration is not at unit, activity would reduce. If it reduces, that means, because if you

check, if you change this, let me, let me put, instead of 1 let me put 10 to the power

minus 2, so I will put E Ni plus plus, Ni would be equal to E 0, which is nothing but

0.24 minus minus because since I have a minus 2 here, so it would be 0.059 into 2. If the

concentration of nickel plus plus would be equal to this unit, so this is my minus 0

point minus 24, so potential will drop. If I increase it, if I again further decrease

the concentration, again this potential will go down. So, that means, you see, as the concentration

of nickel ion concentration, nickel ion concentration is decreasing your potential is also decreasing

for the reaction compared to the standard reduction potential.

So, which means, that nickel ion plus plus concentration, nickel region, if I reduce

this what does it mean? Means, that it means the nickel ion concentration is decreasing.

Or if you consider the other way around, this way around, so nickel has less tendency to

go into the solution because as the concentration is decreasing in the solution, let us say,

I have two beakers, one case concentration is 10 to the power minus 6 nickel plus plus,

concentration is 10 to the power minus 6 where I have a nickel block. And this concentration

is, this is another case I have another nickel block in the water and I have nickel ion concentration

is 10 to the power minus 1. So, it is pretty clear, that here corrosion

is more, here corrosion is less, or if I come from this to this, so I would increase, I

would rather increase the region. So, if I come from this to this what I am doing? I

am basically trying to get into some situation where my corrosion tendency of nickel is going

down because I am going towards a lower, lower concentration. So, now if I place, let us

say, Ni plus plus here and Ni, now this is for one concentration activity or Ni plus

plus equal to 1. Now, if I put 10 to the power minus 2, this

concentration 10 to the power minus 2, this line would go down because here it will go

down. So, it will go down, that means, this would be point for the concentration. For

this if I make it 10 to the power minus 6, then it would be here, this is nickel site.

So, what I am doing? I am extending nickel plus plus region and I am nickel.

Now, let me understand why the nickel is not on top? Nickel plus plus is not on, on the

bottom. Now, coming to this, say an example, let us say I have, I can measure, I can measure

up to the nickel concentration up to 10 to the power minus 6 unit, unit, this can be

molarity or this can be mole fraction. Now, I can measure up to 10 to the power minus

6 and if the concentration of nickel ion is less than 10 to the power minus 6, if the

nickel ion concentration is less than 10 to the power minus 6. So, that means, that if

the concentration, nickel, nickel ion plus plus, nickel ion goes minus 6. That means

that I do not have any existence of nickel ion because I cannot measure below 10 to the

power minus 6 unit of concentration. So, if I cannot measure Ni plus plus, then

what does it mean? That there is no existence of, virtually there is no some nickel plus

plus, so only nickel will remain. If only nickel will remain, if the concentration is

less than 10 to the power 6 or when the measurement limit with the experimental set up, what I

have in that situation, if the measurement limit is 10 to the power minus 6. So, if the

concentration goes below to the minus 6, I would say, that there is no corrosion or there

is no case of Ni plus plus. But if I have this, so there I can say, that there is a

corrosion or nickel, can remain in the presence of nickel, can be present in equilibrium with

Ni plus plus and if it 10 to the power minus 5, then also I will say the corrosion tendency

is increasing or I would say, that the amount of nickel ion concentration is increasing

and nickel is highly getting corroded or the nickel corrosion is substantial. Now, like

that way it will move up. Now, while it is moving up, let us say, in

one case my measurement limit is 10 to the power minus 6, another case my measurement

limit is 10 to the power minus 5, another case measurement limit is 10 to the minus

4. So, as you go from this to this, so you are basically lifting your line upwards or

you are, since this is the measurement limit, so below this, 10 to the minus 6, let us say

this is my measurement limit, 10 to the minus 6. So, below any line I cannot have the existence

of Ni plus plus, this is the measurement limit, 10 to the minus 6. If I have the concentration

10 to the minus 7, then I cannot have any stress of nickel plus plus in this region,

in this region. So, if I do not have nickel plus plus in this region, then there must

be nickel only. Now, my measurement limit, I lift it to 10

to the power minus 2. If I, my, my measurement limit is 10 to the power minus 2, then below

that concentration I will only have Ni because nickel ion concentration is not there. Nickel

ion is not there at all virtually because I cannot measure below to the minus 2. So,

that means, that the lower part of this curve should, should consist of Ni and upper part

should consist of Ni plus plus because this region will tell you, that because of this

factor, because of measurement limit or the limit, what we consider as the percentage

of nickel ion in the solution, that can be measured without experimental data. So, if

I consider in this fashion, then it is very clear, that the lower part should be nickel,

the upper part should be Ni plus plus. Now, we have two regions. Now, the regions

are, this region is Ni plus plus, this region is Ni and of course, I have to specify, that

this line is for concentration Ni plus plus equal to 1. If I go down to the power minus

6, this will also go down, fine. So, now you see, for the first reaction what we have considered

in the beginning that can be put on this E versus pH diagram and it is a horizontal line. So, one part is over, let me see second reaction.

Second reaction was Ni plus 2 H 2 O equal to Ni OH whole 2 plus 2H plus plus 2e. So,

we have to write it in the reverse manner. So, this is, this part would be equal to Ni

plus 2 H 2 O because I need to find out the reduction potential. So, I can also calculate

reduction potential for this process in standard state because I know the value of mu 0 Ni

OH whole 2, I know the value of mu 0 H 2 O, I know the value of mu 0 of Ni and mu 0 of

H plus. This is also a process where electron accepting, electron is getting accepted or

reduction process is taking place. So, I can put it in terms of E, which is now

let me write it in this manner, OH whole 2 H plus or I can say Ox Red would be equal

to E 0 Ox Red plus RT. Here also I have 2 electron, so 2F into ln, Ox is this side because

this side is taking electron, so Ni OH whole 2, the concentration of Ni OH whole 2, concentration

of H plus square because 2 is involved, 2, 2 number of hydrogen ion, ions that are involved

and this is Ni concentration of H 2 O whole square.

Now, this is happening in pure medium. Now, this is a solid, which is precipitating out,

what would be the concentration or activity? This would be one, this is another pure metal,

here also I can consider it as 1. Now, this is also pure H 2 O, which is concentration,

would also be 1. Now, if these are 1, so I can write E 0 Ox by Red plus. Now, I put all

those values, here are T and F at 25 degree Celsius, then this would become 0.059 into,

now we should apply l, Ni would write log H plus square.

Now, if I take this square out there would be a term 2 because 2 is there. So, now, I

can write it as E 0 Ox Red, I can take a minus sign and then I can put the minus sign here

0.059, 2 2 would get cancelled, log of H plus. Let me take this out also, 0.059 minus of

log of H plus, if you see this, what is this, this is nothing but pH of the solution. So,

this would become E 0 Ox Red minus 0.059 pH. Now, this is E 0 Ox Red. Now, this is my equation.

Now, you see this equation has a relation between E 0 that is a standard reduction potential

for this reduction process and pH of the solution. So, this also we can plot in E versus pH and

if you see, that this is also straight line equation because this is a constant and this

is the variable and that is nothing but the pH. So, I can plot this on E versus pH plot

and this should be having a negative slope, which is nothing but minus 0.059. So, I can

also plot it in green color that we plot. So, I can have let me put it in a, this is

nothing but E 0 Ox Red equal to E 0 Ox Red minus 0.059 pH. This is my equation of that

line and this equation was E equal to E 0 plus 0.059 by 2 nickel plus plus. Let me stop here, I can continue in the next lecture.

## 11 comments

## Reza Noori

31:40

How he got 0.059?

2.303*(RT/nF)=2.303*(8.314*298)/(2*96500)=0.029, not 0.059

## Kallol Mondal

Yes, Mr. Reza Noori, the value comes out to be 0.029 if I consider 2. I missed 2 while calculating RT/nF. Thank you very much for pointing this.

## fergure

WOW Sir, you are one amazing educator. THANK YOU.

## AKHIL JAIN

Please correct me if I am wrong since I am unable to find an answer for this.

42:53

At this moment it is mentioned that Standard Electrode Potential is negative and decreasing further when we go from concentration 1 to 10^-6. However, since this is reduction potential, its oxidation potential is increasing (-reduction potential) in the positive direction. So, Ni will have more and more tendency to form Ni++ at 10^-6 Ni++ solution. Hence Ni block will corrode more and more.

## Teja Swaroop Mudiki

Guys any one clarify me how the species/reactions were made 11:09 ….

## Raul Rivera Martinez

It is uppose that 10*-6 M has more tendency to nickel bar corrode than 10*-1M of Ni solution .The teacher has wronged in min 41:16

## Mkuseli Mqana

Guys…could you please tell me how to get the 'mu' values?

## ajad pandey

Sir thodi Hindi bhi bol dijiyy

## Sicaus

thank you!!

## astroid kick

Wow sir thank you very much , salute you 🇮🇳indian profs of iits, my assignment submission was there tomorrow our profs don't teach anything stupid so called NITs

## Debjyoti Mukherjee

অসাধারণ ❤️❤️❤️❤️ আপনাদের জন্যেই বোঝা সম্ভব । engineering knowledge is incomplete without nptel vdeo lectures for noniitans