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:

The cell emf is the reduction potential of the cathode minus that of the anode.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In the reaction, Al is oxidised (anode) and \(\text{Ni}^{2+}\) is reduced (cathode).

Step 2: Identify Electrodes
Cathode: \(E^{\circ}_{Ni} = -0.25\) V. Anode: \(E^{\circ}_{Al} = -1.66\) V.

Step 3: Calculation
\(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.25 - (-1.66) = +1.41\) V.

Step 4: Compare the options
The standard emf is \(+1.41\) V, the third option. The coefficients do not change the value of the potential.

Final Answer:
The standard emf of the cell is plus 1.41 V. \[ \boxed{\text{(C) }+1.41\ \text{V}} \]