A strip of iron is placed in a 1 M solution of iron(II) sulfate, and a strip of

Question

A strip of iron is placed in a 1 M solution of iron(II) 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 to a voltmeter. With the two electrodes connected together, how do the Cl^- ions move? Reduction Half-Reaction E degree(V) Fe^2+ (aq) + 2e^- rightarrow Fe(s) - 0.41 Cu^2+ (aq) + 2e^- rightarr4ow Cu(s) 0.34 A) through the external circuit from Cu to Fe B) through the salt bridge from the Cu half-cell to the Fe half-cell C) in random fashion D) in the direction opposite to the movement of the sulfate ions E) together with the Cu^2+ ions to form an insoluble precipitate A lead storage battery involves the following two half-reactions: PbSO_4(5) + 2e^- rightarrow Pb(s) + SO_4^-2 (aq), E degree = -0.36 V PbO_2(s) + 4H^+(aq) + SO_4^2- (aq) + 2e^- rightarrow PbSO_4(s) + 2H_2O(l); = E degree = 1.69 V During the discharge reaction of the lead storage battery at 1.0 M concentrations, the cell potential and the reducing agent are. respectively. A) -2.05 V and Pb. B) -2.05 V and PbO_2. C) 2.05 V and Pb. D) 2.05 V and PbO_2. E) 1.33 V and Pb. 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? A) Pb is formed at the anode. B) The pH decreases. C) PbO_2 is reduced at the anode. D) SO_4^2- is the reducing agent. E) PbO_2 is the oxidizing agent.

Explanation / Answer

25.B) Through salt bridge from Cu half cell to Fe half cell.

26.C) 2.05V and Pb. The Pb is dissolved in sulphuric acid to form lead lead sulphate and two electron released in this half reaction and this two electron used to reduce PbO2 to PbSO4 in cathodic half reaction.

27.E. PbO2 is Oxidising agent.PbO2 is itself reduced to PbSO4 and it oxize Pb to PbSO4.

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