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Q.a) Write the equations for the reactions taking place at anode and cathode in the Lead-storage battery.

(3)
b) Calculate the value of ΔrG° at 298 K for the cell reaction.
3Mg(s) + 2Al³⁺(aq) → 3Mg²⁺(aq) + 2Al(s)
[Given ; E°_Mg = –2.36 V, E°_Al = –1.66 V and F = 96487 C]. (2)
Karnataka PUCKarnataka II PUC Board 2019Subjective· 5mImportance★★★★★
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Lead battery: Pb anode and PbO2_2 cathode both form PbSO4_4 in H2_2SO4_4. For the Mg–Al cell, Ecell∘=0.70E^\circ_{cell} = 0.70 V, n=6n=6, so ΔrG∘=−nFE∘=−405.25\Delta_r G^\circ = -nFE^\circ = -405.25 kJ.

a) Lead storage battery (on discharge):

  • Anode (oxidation): Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-
  • Cathode (reduction): PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l)PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)
  • Overall cell reaction: Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)

b) Calculation of ΔrG∘\Delta_r G^\circ:

Reaction: 3Mg(s)+2Al3+(aq)→3Mg2+(aq)+2Al(s)3Mg(s) + 2Al^{3+}(aq) \rightarrow 3Mg^{2+}(aq) + 2Al(s).

Here Mg is oxidised (anode) and Al3+^{3+} is reduced (cathode).

Cell emf:

Ecell∘=Ecathode∘−Eanode∘=EAl3+/Al∘−EMg2+/Mg∘=(−1.66)−(−2.36)=0.70 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Al^{3+}/Al} - E^\circ_{Mg^{2+}/Mg} = (-1.66) - (-2.36) = 0.70\ \text{V}

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