Question

A platinum electrode is immersed in a solution containing \[0.1M\,F{e^{2 + }}\] and \[0.1M\,F{e^{3 + }}\]. It is coupled with HER. Concentration of \[F{e^{3 + }}\] is increased to \[1M\] without change in \[\left[ {F{e^{2 + }}} \right]\] then change in EMF (in centivolt) is:
$ \left( A \right)0.6 \\
  \left( B \right)60 \\
  \left( C \right)0.06 \\
  \left( D \right)6 \\
$

Answer 1

Hint: In order to solve this question, we are going to first see the concentration of the \[F{e^{3 + }}\] and the \[F{e^{2 + }}\] ions and find the initial EMF of the cell from that. After this, we are going to see the final concentrations of both the ions and find the EMF then, then converting the EMF to centivolt, we get the answer.

Formula used: The EMF of a cell is given by
 \[{E_{cell}} = {E^0}_{cell} - \dfrac{{0.059}}{n}log({K_C})\]
Where, \[n\] is the number of the moles present in the solution and \[{K_C}\] is the ratio of concentration of the two types of ions.

Complete step by step solution:
It is given in this question that the concentration of the \[\,F{e^{2 + }}\] and \[F{e^{3 + }}\] are:
$
  \left[ {F{e^{2 + }}} \right] = 0.1M \\
  \left[ {F{e^{3 + }}} \right] = 0.1M \\
$
At this point, the EMF of the cell is calculated as:

$
  {E_{cell}} = 0.77 - \dfrac{{.059}}{1}\log \dfrac{{0.1}}{{0.1}} \\
   \Rightarrow {E_{cell}} = 0.77V \\
$
Now, after coupling with SHE, the concentration of\[F{e^{3 + }}\] increases to \[1M\] and the concentration of \[\left[ {F{e^{2 + }}} \right]\] remains the same,
Therefore, the EMF is calculated as:
$
 {E_{cell}}' = 0.77 - \dfrac{{0.059}}{1}\log \dfrac{{0.1}}{1} \\
   \Rightarrow {E_{cell}}' = 0.829V \\
$
Thus, the change in the EMFs before and after is
\[{E_{cell}}' - {E_{cell}} = 0.829V - 0.77V = 0.059V\]
Thus, in centivolt the change in the EMF becomes, \[6cV\]

Hence, option \[\left( D \right)6\] is the correct answer.

Note: The EMF of a cell depends upon the initial EMF of the cell which is also denoted as \[{E^0}_{cell}\] present when there is no potential applied across the cell. For the solution containing the \[F{e^{3 + }}\] and the \[F{e^{2 + }}\] ions, the value of the initial EMF of the cell, \[{E^0}_{cell}\] is \[0.77V\] and this fact has been seen in this question.