Question
Question: A platinum electrode is immersed in a solution containing \[0.1M\,F{e^{2 + }}\] and \[0.1M\,F{e^{3 +...
A platinum electrode is immersed in a solution containing 0.1MFe2+ and 0.1MFe3+. It is coupled with HER. Concentration of Fe3+ is increased to 1M without change in[Fe2+]then change in EMF (in centivolt) is:
(A)0.6 (B)60 (C)0.06 (D)6
Solution
In order to solve this question, we are going to first see the concentration of the Fe3+ and the Fe2+ 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
Ecell=E0cell−n0.059log(KC)
Where, n is the number of the moles present in the solution and KC is the ratio of concentration of the two types of ions.
Complete step by step answer:
It is given in this question that the concentration of the Fe2+ and Fe3+ are:
[Fe2+]=0.1M [Fe3+]=0.1M
At this point, the EMF of the cell is calculated as:
Ecell=0.77−1.059log0.10.1 ⇒Ecell=0.77V
Now, after coupling with SHE, the concentration of Fe3+ increases to 1M and the concentration of [Fe2+] remains the same,
Therefore, the EMF is calculated as:
Ecell′=0.77−10.059log10.1 ⇒Ecell′=0.829V
Thus, the change in the EMFs before and after is
Ecell′−Ecell=0.829V−0.77V=0.059V
Thus, in centivolt the change in the EMF becomes, 6cV
Hence, option (D) is the correct answer.
Note: The EMF of a cell depends upon the initial EMF of the cell which is also denoted as E0cell present when there is no potential applied across the cell. For the solution containing the Fe3+ and theFe2+ ions, the value of the initial EMF of the cell, E0cell is 0.77V and this fact has been seen in this question.