Chapter 9: Problem 117
In electrolysis of \(\mathrm{NaCl}\) when \(\mathrm{Pt}\) electrode is taken then \(\mathrm{H}_{2}\) is liberated at cathode while with \(\mathrm{Hg}\) cathode, it forms sodium amalgam. The reason for this is: (a) more voltage is required to reduce \(\mathrm{H}^{+}\) at \(\mathrm{Hg}\) than at Pt. (b) concentration of \(\mathrm{H}^{+}\) ions is larger when \(\mathrm{Pt}\) ele trode is taken. (c) Na is dissolved in \(\mathrm{Hg}\) while it does not dissolve in Pt. (d) \(\mathrm{Hg}\) is more inert than \(\mathrm{Pt}\).
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.