Question

If the standard reduction potential (\( E^\circ \)) of \( \left( \text{Ni}^{2+}_{\text{(aq)}} \mid \text{Ni}_{\text{(s)}} \right) \) and \( \left( \text{Al}^{3+}_{\text{(aq)}} \mid \text{Al}_{\text{(s)}} \right) \) are \( -0.25 \, \text{V} \) and \( -1.66 \, \text{V} \) respectively, then what is the standard emf of the cell reaction? 
 
\[ 2\text{Al}_{\text{(s)}} + 3\text{Ni}^{2+}_{\text{(aq)}} \longrightarrow 2\text{Al}^{3+}_{\text{(aq)}} + 3\text{Ni}_{\text{(s)}} \]

• A. +2.57 V
• B. -2.57 V
• C. +1.41 V
• D. -1.91 V

Hint:

Normally use: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] If the option key differs, keep the chemistry calculation noted separately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The standard EMF of a cell (\(E^\circ_{\text{cell}}\)) is calculated using the standard reduction potentials of the cathode and anode. The species that is reduced acts as the cathode, and the species that is oxidized acts as the anode.
Step 2: Key Formula or Approach:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Where both values are standard reduction potentials.
Step 3: Detailed Explanation:
From the cell reaction:
\[ 2\text{Al}_{\text{(s)}} + 3\text{Ni}^{+2}_{\text{(aq)}} \longrightarrow 2\text{Al}^{+3}_{\text{(aq)}} + 3\text{Ni}_{\text{(s)}} \] - Aluminum (\(\text{Al}\)) is being oxidized: \(\text{Al} \rightarrow \text{Al}^{+3} + 3\text{e}^-\) (Anode)
- Nickel (\(\text{Ni}^{+2}\)) is being reduced: \(\text{Ni}^{+2} + 2\text{e}^- \rightarrow \text{Ni}\) (Cathode)
Given:
\(E^\circ_{\text{cathode}} (E^\circ_{\text{Ni}^{+2}/\text{Ni}}) = -0.25 \text{ V}\)
\(E^\circ_{\text{anode}} (E^\circ_{\text{Al}^{+3}/\text{Al}}) = -1.66 \text{ V}\)
Calculation:
\[ E^\circ_{\text{cell}} = -0.25 \text{ V} - (-1.66 \text{ V}) \] \[ E^\circ_{\text{cell}} = -0.25 + 1.66 = +1.41 \text{ V} \] Step 4: Final Answer:
The standard emf is +1.41 V.