Question

The \(E_{\text{cell}}^0\) of \(\text{Al}_{(s)}|\text{Al}^{3+}(1\text{M})||\text{Pb}^{2+}(1\text{M})|\text{Pb}_{(s)}\) cell is 1.5 V if \(E_{\text{Pb}}^0\) is \(-0.14\) V then \(E_{\text{Al}}^0\) will be

• A. \(1.64\) V
• B. \(-1.64\) V
• C. \(1.36\) V
• D. \(-1.36\) V

Hint:

E cell = E cathode - E anode; lead is the cathode and aluminium the anode.

Answer 1

The Correct Option is B

Step 1: Identify electrodes
Left electrode Al is oxidised, so it is the anode; right electrode Pb is reduced, so it is the cathode.

Step 2: Equation
$E^0_{cell} = E^0_{Pb} - E^0_{Al}$, so $1.5 = -0.14 - E^0_{Al}$.

Step 3: Solve
$E^0_{Al} = -1.64$ V. Option (B).

Final Answer:
Option (B). \[ \boxed{-1.64 \text{ V}} \]