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Question 93 of 111

Q.Write the cell reaction and calculate Ecell∘E^\circ_{cell} of the following electrochemical cell: Al(s) ∣ Al3+(aq.)(1M) ∣∣ Zn2+(aq.)(1M) ∣ Zn(s)Al(s)\,|\,Al^{3+}(aq.)(1M)\,||\,Zn^{2+}(aq.)(1M)\,|\,Zn(s); EAl∘=−1.66 VE^\circ_{Al} = -1.66\ V, EZn∘=−0.76 VE^\circ_{Zn} = -0.76\ V

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 2mImportance★★★★★
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Al (more negative E∘E^\circ) is oxidised at the anode, Zn2+Zn^{2+} is reduced at the cathode; Ecell∘=+0.90 VE^\circ_{cell} = +0.90\ V.

In Al(s) ∣ Al3+(aq)(1M) ∣∣ Zn2+(aq)(1M) ∣ Zn(s)Al(s)\,|\,Al^{3+}(aq)(1M)\,||\,Zn^{2+}(aq)(1M)\,|\,Zn(s), aluminium (left electrode, more negative E∘E^\circ) is the anode where oxidation occurs, and zinc is the cathode where reduction occurs. Balancing electrons (Al loses 3e⁻, Zn gains 2e⁻; LCM = 6):

  • Anode: 2Al→2Al3++6e−2Al \rightarrow 2Al^{3+} + 6e^-
  • Cathode: 3Zn2++6e−→3Zn3Zn^{2+} + 6e^- \rightarrow 3Zn …

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