91Ó°ÊÓ

• Electrode potential (E) is the electromotive force that exists between two electrodes. Two electrodes make up a cell: one is a standard electrode (such as calomel electrode or standard hydrogen electrode) and the other is a provided electrode.
• A titration is "the process of determining the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point")," according to the definition.
• The existence or change in concentration of the oxidised and reduced forms of a redox pair is sensed by an indicator electrode in redox techniques. The indicator electrode is usually made of an inert noble metal, such as Pt, and the cell's potential is measured against a reference electrode.

(a)

Take a look at the titration in the diagram; a balanced titration reaction must be written.

In a chemical reaction, there is a mass conversion law that states that the total mass of the product must equal the total mass of the reactants.

The conversion of reactants into products is central to the concept of writing a balanced chemical reaction.

First, write a reaction based on the facts you've been provided.

Then, in both reactants and products, count the number of atoms of each element.

Finally, the acquired values might be classified as reactant and product coefficients.

The titration of iron (II) with cerium is shown in Figure 1. (IV)\(100.0\;mL of0.0500MF{e^{2 + }}\)\(with0.100MCeC{e^4}\)in\(1MHCl{O_4}\). Each Ceric ion participates in this process \(\left( {C{e^{4 + }}} \right)\)regularly lowers one mole of ferrous ion As a result, a balanced titration response is achieved,

\(C{e^{4 + }} + F{e^{2 + }} \to C{e^{3 + }} + F{e^{3 + }}\)

(b)

Two separate half-reactions for the indicator electrode must be written for the titration shown in the illustration.

Introduction to the concept:

Half reaction: The part of the reaction in which the oxidation or reduction reaction (redox reaction) occurs. A change in the oxidation state of a specific material participating in a redox reaction produces a half-reaction.

This indicator's half-reaction time is,

There are two processes that reach equilibrium in the Pt indicator electrode

Ferrous is converted to ferric, and cerium is converted to ceric.

(c)

To write two distinct Nernst equations for cell voltage, consider the titration shown in the diagram.

The Nernst equation is as follows:
\(E = {E_0} - \frac{{RT}}{{nF}}lnQ\)

Where,

\({E_0} - \)Standard potential of the cell

R - Gas constant (8.314J/K.Mol)

T - Absolute temperature (Kelvin)

n - Number of electrons

F- Faraday's constant (\(96485C/\)mole)

Q - Reactant quotient

The cell voltage is calculated using the Nernst equation.

The following are the two indication reactions:

For these reactions, we should use the Nernst equation.

So,

\(E = \left\{ {0.767 - 0.05916log\frac{{\left| {F{e^{2 + }}} \right|}}{{\left| {F{e^{1 + }}} \right|}}} \right\} - \{ 0.241\} \)

Formal prospect for lowering\(F{e^{3 + }}\)in\(1MHCl{O_4}\)is\(0.767\;V\).

A saturated calomel electrode has a potential of\(0.241\;V\).

\(E = \left\{ {1.70 - 0.059\mid 6log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{4 + }}} \right|}}} \right\} - \{ 0.24\mid \} \)

TheNernst equations for the cell voltage

\(\begin{aligned}{l}E &= \left\{ {0.767 - 0.05916log\frac{{\left| {F{e^{2 + }}} \right|}}{{\left| {F{e^3}} \right|}}} \right\} - \{ 0.241\} \\E &= \left\{ {1.70 - 0.05916log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{44}}} \right|}}} \right\} - \{ 0.241)\end{aligned}\)

(d)

In the case of the titration shown in the diagram, the value of E must be computed for the given volume.

\(E = {E_ + } + {E_ - }\)

Where,

\({E_ + }\)is the electrode's potential when it's linked to the positive terminal

\({E_ - }\)is the electrode's potential when it's linked to the negative terminal

Titration:

\(C{e^{4 + }} + F{e^{2 + }} \to C{e^{3 + }} + F{e^{3 + }}\)

As each component of\(C{e^{4 + }}\)when it is added, it consumes\(C{e^{4 + }}\)and produces the same number of moles of\(C{e^{3 + }}\)and\(F{e^{3 + }}\). More unreacted material was present prior to the equivalence point \(F{e^{2 + }}\)rests in the aqueous solution

At 50.0mL, the equivalence point is reached.

Prior to the moment of equivalency,

\(\begin{aligned}{c}E &= {E_ + } + {E_ - }\\ &= {E_ + } + E(calomal)\\ &= \left( {0.767 - 0.05916log\frac{{\left| {F{e^{2 + }}} \right|}}{{\left| {F{e^2}} \right|}}} \right) - 0.241\\\left( {0.526 - 0.05916log\frac{{\left| {F{E^{2 + }}} \right|}}{{\left( {F{e^2}\mid } \right.}}} \right)......(1)\end{aligned}\)

At\(10.0\;mL:\)

This is the equivalence point method. As a result, 10.0/50.0 percent of the iron is in the form of\(F{e^{3 + }}\)thus the form of 40.0 / 50.0 is\(F{e^{2 + }}\). Using this value as a substitute in equation (1),

\(\begin{aligned}{l} &= \left( {0.526 - 0.05916log\frac{{(4a50)}}{{(1050)}}} \right)\\ &= 0.526 - 0.05916log(4.0)\\ &= 0.526 - 0.05916(0.6020)\\ &= 0.526 - (0.0356)\\ &= 0.490\;V\end{aligned}\)

At\(25.0\;mL\):

25.0 / 50.0 percent of iron is in the form of\(F{e^{3 + }}\)thus the form of 25.0 / 50.0 is\(F{e^{2 + }}\) Using this value as a substitute in equation (1),

\(\begin{aligned}{l} &= \left( {0.526 - 0.05916log\frac{{(25/50)}}{{(25/50)}}} \right)\\ &= 0.526 - 0.05916log(1)\\ &= 0.526 - 0.05916(0)\\ &= 0.526\;V\end{aligned}\)

At\(49.0\;mL\):

Iron in the form of 49.0/50.0 percent\(F{e^{3 + }}\)where the form of 1.0 / 50.0 is\(F{e^{2 + }}\). Using this value as a substitute in equation (1),

\(\begin{aligned}{l} &= \left( {0.767 - 0.05916log\frac{{(1.050)}}{{(49.0050)}}} \right) - 0.241\\ &= (0.767 - 0.05916log(0.0204)) - 0.241\\ &= (0.767 - 0.05916( - 1.6903)) - 0.241\\ &= (0.767 - ( - 0.1000)) - 0.241\\ &= 0.867 - 0.241\\ &= 0.626\;V\end{aligned}\)

At\(50.0\;mL\)

This is the point of equivalency\(\left( {{V_e}} \right)\), where\(\left( {C{e^{3 + }}} \right)\)is the same as\(\left( {F{e^{3 + }}} \right)\)and\(\left( {C{e^{4 + }}} \right)\) is the same as\(F{e^{2 + }}\). The Nernst equation is used in this reaction,

\(\begin{aligned}{l}{E_4} &= 0.767 - 0.05916log\frac{{\left| {F{e^{3 + }}} \right|}}{{\left| {F{e^{3 + }}} \right|}}........(1)\\{E_ + } &= 1.70 - 0.05916log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{4 + }}} \right|}}..........(2)\end{aligned}\)

By combining these two voltages,

\(2{E_ + } = 2.467 - 0.05916log\left( {\frac{{\left| {F{e^{2 + }}} \right|\left| {C{e^{2 + }}} \right|}}{{\left| {F{e^{2 + }}} \right|\left| {C{e^{4 + }}} \right|}}} \right)\)

\(\left( {C{e^{3 + }}} \right) = \left( {F{e^{3 + }}} \right)\)and

\(\left( {C{e^{4 + }}} \right) = \left( {F{e^{2 + }}} \right)\),

The concentration ratio is unity at the equivalent point. As a result, the logarithm is zero.

\(\begin{aligned}{l}2{E_4} = 2.467\;V\\{E_ + } = 1.23\;V\end{aligned}\)

The voltage in the cell is,

\(\begin{aligned}{c}E = {E_ + } - E (Calomel) \\ = 1.23 - 0.241\\ = 0.99\;V\end{aligned}\)

At\(51.0\;mL\):

Following the equivalence point, the formula is used to compute the cell's voltage,

\(\begin{aligned}{l}E &= \left( {1.70 - 0.05916log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{4 + }}} \right|}}} \right) - 0.241\\ &= \left( {1.70 - 0.05916log\frac{{50.0}}{{1.0}}} \right) - 0.241\\ &= (1.70 - 0.05916 \times (1.6989)) - 0.241\\ &= (1.70 - 0.1005) - 0.241\\ &= - 1.5995 - 0.241\\ &= 1.36\;V\end{aligned}\)

\(\begin{aligned}{l}At 60.0\;mL : \\E &= \left( {1.70 - 0.05916log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{4 + }}} \right|}}} \right) - 0.241\\ &= \left( {1.70 - 0.05916log\frac{{50.0}}{{10.0}}} \right) - 0.241\\ &= (1.70 - 0.05916 \times (0.6989)) - 0.241\\ &= (1.70 - 0.0413) - 0.241\\ &= 1.6587 - 0.241\\ &= 1.42\;V\\ &= 1.36\;V \end{aligned}\)

\(\begin{aligned}{l}At 100.0\;mL :\\ E &= \left( {1.70 - 0.05916log\frac{{\left| {C{e^{3 + }}} \right|}}{{\left| {C{e^{4 + }}} \right|}}} \right) - 0.241\\ &= \left( {1.70 - 0.05916log\frac{{50.0}}{{50.0}}} \right) - 0.241\\ &= (1.70 - (0)) - 0.241\\ &= 1.46\;V\end{aligned}\)