Question

How to do redox titration calculation?

Answer 1

Hint: Redox titration is a type of titration in which oxidation and reduction takes place and the change in electrode potential is seen. Calculations for redox titration are solved using graphs. The graph is made in between change in potential versus concentration of ${M^{n + }}$ here it is $pM$ similar to $pH$ .

Complete solution:
Oxidation-reduction titration which is also known as redox titration is used as an analytical technique, where one reactant reduces while other oxidizes. Their change in oxidation number or phenomenon of oxidation or reduction is generally seen by the change in color. Mostly redox reaction is seen between $F{e^{ + 2}}$ and $C{e^{ + 4}}$ during the redox titration, $F{e^{ + 2}}$ changes to $F{e^{ + 3}}$ by oxidation or we can also say by loss of electrons. Similarly, $C{e^{ + 4}}$ changes to $C{e^{ + 3}}$ by reduction or we say it as gain of electrons as we termed for iron.
For starting the calculation, firstly we have to understand the basics. Here a graph which is called a titration curve is plotted between the potential change and the concentration change. We know from Nernst equation, if we apply it on any ionic solution like for $F{e^{ + 2}}$ and $C{e^{ + 4}}$ , for this redox reaction if we write Nernst equation it looks like this-
$F{e^{ + 2}} + C{e^{ + 4}} \rightleftharpoons \,F{e^{ + 3}} + C{e^{ + 3}}$
For iron ${E_{cell}} = E_{cell\left( {F{e^{ + 3}},F{e^{ + 2}}} \right)}^ \circ - \dfrac{{RT}}{{nF}}\ln \left( {\dfrac{{F{e^{ + 3}}}}{{F{e^{ + 2}}}}} \right)$ and
For cerium it will be ${E_{cell}} = E_{cell\left( {C{e^{ + 4}},C{e^{ + 3}}} \right)}^ \circ - \dfrac{{RT}}{{nF}}\ln \left( {\dfrac{{C{e^{ + 3}}}}{{C{e^{ + 4}}}}} \right)$
Where, $E_{cell}^ \circ = E_{cathode}^ \circ - E_{anode}^ \circ $
Before the equivalence point, it is easy for us to calculate the amount of unreacted $F{e^{ + 2}}$ by using Nernst equation for Iron. Similarly after the equivalence point, it is easy to calculate concentration of unreacted $C{e^{ + 4}}$ by using its Nernst equation.
Now at point of equivalence we have to calculate the average of two ions $F{e^{ + 2}}$ and $C{e^{ + 4}}$ because at that time the concentration of both ions and their oxidised and reduced version become equal.
${E_{eq}} = \,E_{cell\left( {F{e^{ + 3}},F{e^{ + 2}}} \right)}^ \circ + \,E_{cell\left( {C{e^{ + 4}},C{e^{ + 3}}} \right)}^ \circ $ So, we have to just calculate $E_{cell}^ \circ $ for both the reactions and then taking average of both.
Like this we can attempt calculation of redox titration.

Note:The titrations are used to find the potential difference when a certain reaction is going on in the container. Calculate average of $E_{cell}^ \circ $ for both the reactions only at the point of equivalence. Before and after equivalence point we firstly have to see which reaction among $\left( {C{e^{ + 4}},C{e^{ + 3}}} \right)$ or $\left( {F{e^{ + 3}},F{e^{ + 2}}} \right)$ is going on.