TS EAMCET · Chemistry · Electrochemistry
The alkali metal with lowest \(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}(\mathrm{V})\) is \(\mathrm{X}\) and the alkali metal with highest \(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\circ}(\mathrm{V})\) is \(\mathrm{Y}\). Then \(\mathrm{X}\) and Y are respectively
- A \(\mathrm{Li}, \mathrm{Na}\)
- B \(\mathrm{Li}, \mathrm{Cs}\)
- C \(\mathrm{Na}, \mathrm{Li}\)
- D \(\mathrm{Cs}, \mathrm{Li}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{Li}, \mathrm{Na}\)
Step-by-step Solution
Detailed explanation
The standard reduction potential of \(\mathrm{Li}^{+}\)is lowest because of its exceptionally small size and high chargeto-mass ratio. It is highest for \(\mathrm{Na}^{+}\)to \(\mathrm{Na}\) due to its size, electropositivity and hydration enthalpy.
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