Chemistry MCQs for NEET — Practice Questions with Answers

The Nernst equation for the given reaction is: E = E$^{\text{o}}$ - RT/nF ln(1/[M$^{n+}$]). If the concentration [M$^{n+}$] decreases, then 1/[M$^{n+}$] increases. Since the term -RT/nF ln(1/[M$^{n+}$]) is subtracted from E$^{\text{o}}$, an increase in ln(1/[M$^{n+}$]) will lead to a more negative value being subtracted, thus resulting in a decrease in the electrode potential (E). It will become more negative or less positive.

The text states, 'The standard potential of the cell can be obtained by taking the difference of the standard potentials of cathode and anode (E$^{\text{o}}_{\text{cell}}$ = E$^{\text{o}}_{\text{cathode}}$ – E$^{\text{o}}_{\text{anode}}$)'. For a spontaneous reaction, E$^{\text{o}}_{\text{cell}}$ must be positive, and $\Delta_{\text{r}}G^{\text{o}}$ must be negative ($\Delta_{\text{r}}G^{\text{o}}$ = -nFE$^{\text{o}}_{\text{cell}}$).

A negative E$^{\text{o}}$ for the reduction of V$^{3+}$ to V$^{2+}$ (-0.26 V) implies that V$^{3+}$ is not easily reduced, or conversely, V$^{2+}$ is easily oxidized. However, the text mentions 'The comparatively low value for V is related to the stability of V$^{2+}$ (half-filled t$_{2g}$ level)'. A negative E$^{\text{o}}$ means the reverse reaction (oxidation of V$^{2+}$ to V$^{3+}$) is relatively favorable. The stability of V$^{2+}$ would make it reluctant to get oxidized. Re-evaluating based on the question and context, a negative E$^{\text{o}}$ means V$^{3+}$ is actually harder to reduce than H$^+$, and therefore not a strong oxidizing agent. Its reverse reaction is V$^{2+}$ $\to$ V$^{3+}$ + e$^{-}$ with E$^{\text{o}}$ = +0.26 V. This positive potential indicates that V$^{2+}$ would readily get oxidized, meaning V$^{2+}$ is a good reducing agent, or it is not particularly stable and tends to lose an electron. However, the context says that the 'low value' (negative) for V (referring to M$^{3+}$/M$^{2+}$) is because of the stability of V$^{2+}$. So option 4 aligns with the direct interpretation of the text, meaning V$^{2+}$ is relatively stable due to electron configuration, hence its tendency to get oxidized is lower than expected from a simpler trend, leading to a negative potential for V$^{3+}$ to V$^{2+}$ reduction.

The summary explicitly states: 'The standard potential of the cells are related to standard Gibbs energy ($\Delta_{\text{r}}G^{\text{o}}$ = – nF E$^{\text{o}}_{\text{cell}}$) and equilibrium constant ($\Delta_{\text{r}}G^{\text{o}}$ = – R T ln K) of the reaction taking place in the cell.'

The text explains: 'Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in the reaction but provide their surface for oxidation or reduction reactions and for the conduction of electrons.'

The text states, 'The low value for Sc (referring to E$^{\text{o}}$ (M$^{3+}$/M$^{2+}$) for elements that can form M$^{2+}$) reflects the stability of Sc$^{3+}$ which has a noble gas configuration.' Sc$^{3+}$ is extremely stable and difficult to reduce to a lower oxidation state (or metal), making the standard potential for Sc$^{3+}$/Sc irrelevant in the context of standard M$^{2+}$/M potentials or implying that Sc$^{2+}$ is not readily formed or reduced.

A positive E$^{\text{o}}$ for a reduction half-reaction (e.g., Ag$^{+}$ + e$^{-}$ $\to$ Ag, E$^{\text{o}}$ = +0.80V) indicates that the species being reduced (Ag$^{+}$ in this case) has a strong tendency to gain electrons and get reduced. A species that readily accepts electrons and gets reduced is a strong oxidizing agent.

According to the NCERT text, 'In 1800, only 31 elements were known. By 1865, the number of identified elements had more than doubled to 63.' This directly supports option o1.

The NCERT text states, 'The German chemist, Johann Dobereiner in early 1800’s was the first to consider the idea of trends among properties of elements. By 1829 he noted a similarity among the physical and chemical properties of several groups of three elements (triads).'

The NCERT text explains Dobereiner's observation: 'In each case, he noticed that the middle element of each of the triads had an atomic weight about half way between the atomic weights of the other two (Table 3.1).'