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
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
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E cell = E cathode - E anode; lead is the cathode and aluminium the anode.
Updated On: Oct 1, 2026
- \(1.64\) V
- \(-1.64\) V
- \(1.36\) V
- \(-1.36\) V
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
In the cell notation anode then cathode, the left electrode is the anode (oxidation) and the right one is the cathode (reduction). \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\), using reduction potentials.
Step 2: Detailed Explanation
Here Al is the anode and Pb is the cathode.
\[ 1.5 = (-0.14) - E^0_{Al} \]
\[ E^0_{Al} = -0.14 - 1.5 = -1.64 \text{ V} \]
A positive value like +1.64 V (A) would be the oxidation potential, not the reduction potential asked for. Aluminium is more reactive than lead, so a negative reduction potential is expected.
Final Answer:
\(E^0_{Al} = -1.64\) V, option (B). \[ \boxed{-1.64 \text{ V}} \]
In the cell notation anode then cathode, the left electrode is the anode (oxidation) and the right one is the cathode (reduction). \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\), using reduction potentials.
Step 2: Detailed Explanation
Here Al is the anode and Pb is the cathode.
\[ 1.5 = (-0.14) - E^0_{Al} \]
\[ E^0_{Al} = -0.14 - 1.5 = -1.64 \text{ V} \]
A positive value like +1.64 V (A) would be the oxidation potential, not the reduction potential asked for. Aluminium is more reactive than lead, so a negative reduction potential is expected.
Final Answer:
\(E^0_{Al} = -1.64\) V, option (B). \[ \boxed{-1.64 \text{ V}} \]
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