A galvanic cell is composed of two hydrogen electrodes, of which one is a standard one. In which of the following solution should the other electrode be immersed to get maximum emf
Answer & explanation
Correct answer: option 2
The correct answer is option 2. \(0.1\, \ M\, \ CH_3COOH\).
The emf (electromotive force) of a galvanic cell can be calculated using the Nernst equation for hydrogen electrodes. When comparing a standard hydrogen electrode (SHE) with another hydrogen electrode immersed in a solution of interest, the difference in hydrogen ion concentration (\([H^+]\)) between the two electrodes determines the cell potential.
Given:
One electrode is a standard hydrogen electrode (SHE), which is in a 1 M \([H^+]\) solution with a potential of 0 V. The other electrode is immersed in a solution of the options provided.
The Nernst equation for the hydrogen electrode can be written as:
\( E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[H^+]}{1} \right) \)
where \( E^\circ \) is the standard electrode potential (0 V for the SHE), \( n \) is the number of electrons transferred (2 for the hydrogen electrode), and \([H^+]\) is the hydrogen ion concentration in the solution.
The given solutions are:
1. 0.1 M HCl (strong acid, fully dissociated): Fully dissociated, \([H^+]\) = 0.1 M
2. 0.1 M \(CH_3COOH\) (weak acid, partially dissociated): Partially dissociated, \([H^+]\) is much less than 0.1 M
3. 0.1 M \(H_3PO_4\) (weak acid, partially dissociated): Partially dissociated, \([H^+]\) is less than 0.1 M
4. 0.1 M \(H_2SO_4\) (strong acid, fully dissociated): Fully dissociated in the first dissociation step, \([H^+]\) = 0.1 M
The maximum emf will be obtained when the difference in \([H^+]\) between the two electrodes is the greatest.
Since \(CH_3COOH\) is a weak acid and only partially dissociates, the concentration of hydrogen ions in a 0.1 M solution of acetic acid is much lower than 0.1 M, leading to a lower hydrogen ion concentration compared to the solutions of strong acids (HCl and \(H_2SO_4\)).
Given the lower \([H^+]\) in the acetic acid solution, the difference in \([H^+]\) between the standard hydrogen electrode and the electrode in the acetic acid solution will be the greatest, resulting in a higher emf according to the Nernst equation.
Thus, the correct answer is: 0.1 M \(CH_3COOH\)