Question

The standard electrode potential (in V) values for Al$^{3+}$/Al, Tl$^{3+}$/Tl are respectively

• A. -1.66, -1.26
• B. +1.66, +1.26
• C. -1.66, +1.26
• D. +1.66,-1.26

Hint:

Standard electrode potentials are always given as reduction potentials. A more negative value indicates a greater tendency to be oxidized, while a more positive value indicates a greater tendency to be reduced. When combining half-reactions, remember that the potential is an intensive property and does not depend on the stoichiometric coefficients.

Answer 1

The Correct Option is C

Step 1: Identify the required standard electrode potentials.
We need the standard reduction potentials for the following half-reactions: $$\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \quad E^\circ(\text{Al}^{3+}/\text{Al})$$ $$\text{Tl}^{3+}(aq) + 2e^- \rightarrow \text{Tl}^{+}(aq) \quad E^\circ(\text{Tl}^{3+}/\text{Tl}^{+})$$ $$\text{Tl}^{+}(aq) + e^- \rightarrow \text{Tl}(s) \quad E^\circ(\text{Tl}^{+}/\text{Tl})$$ The question asks for Tl$^{3+}$/Tl, which involves a direct reduction of Tl$^{3+}$ to Tl. We need to combine the steps for Tl$^{3+}$ to Tl$^{+}$ and Tl$^{+}$ to Tl. However, the provided options directly give values for Al$^{3+}$/Al and Tl$^{3+}$/Tl. We will rely on standard electrochemical data.
Step 2: Look up the standard reduction potentials.
From standard electrochemical series:
$$E^\circ(\text{Al}^{3+}/\text{Al}) = -1.66 \text{ V}$$
$$E^\circ(\text{Tl}^{3+}/\text{Tl}) = +1.26 \text{ V}$$

Step 3: Match the values with the options.
The standard electrode potentials for Al$^{3+}$/Al and Tl$^{3+}$/Tl are -1.66 V and +1.26 V, respectively. This matches option 3. Final Answer: The final answer is $\boxed{-1.66, +1.26}$