The overall reaction for the lead storage battery is as follows: Pb(s) + PbO_2(s

Question

The overall reaction for the lead storage battery is as follows: Pb(s) + PbO_2(s) + 4H^+(aq) + 2SO_4^2-(aq) rightarrow 2PbSO_4(s) + 2H_2O(l) When the battery discharges, which of the following statements is true? Pb is formed at the anode. The pH decreases. PbO_2 is reduced at the anode. SO_4^2- is the reducing agent. PbO_2 is the oxidizing agent. A strip of iron is placed in a 1 M solution of iron(ll) sulfate, and a strip of copper is placed in a 1 M solution of copper(II) chloride. The two solutions are connected with a salt bridge, and the two metals are connected by a wire. Reduction Half-Reaction E degree (V) Fe^2+(aq) + 2e^- rightarrow Fe(s) -0.41 Cu^2+(aq) + 2e^- rightarrow Cu(s) 0.34 Which of the following takes place? The Fe(II) concentration of the iron half-cell decreases Chlorine is produced at the iron electrode. Chlorine is produced at the copper electrode. Copper atoms deposit at the cathode. Sulfur deposits at the iron electrode.

Explanation / Answer

27)

The reaction :- Pb (s) + PbO2 (s) + 2H+ (aq) + 2SO42-  (aq) = 2PbSO4 (s) + 2H2O (l),

When the cell discharges the reaction proceeds to the right, Here PbO2 reduced to PbSO4 at cathode and it oxidises the Pb (s) at anode. So PbO2 acts as oxidizing agent is the true statement.

28) From the reduction elctrode potential values, we can say that feasible cell will show positive potential only when the Cu2+ to Cu(s) occurs at right hand side and Fe(s) to Fe2+ occurs at left hand side. In electrochemical cell, reduction happens at cathode. So correct answer is copper atom is deposited at cathode.

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