Final answer:
In this battery, the copper electrode is the anode, and the lead electrode is the cathode. The overall reaction is 2Cu(s) + Pb²+(aq) -> 2Cu²+(aq) + Pb(s). The standard cell potential for this battery is -0.47 V. As the battery operates, the concentration of Cu²+ ions in the copper half-cell will decrease, while the concentration of Pb²+ ions in the lead half-cell will increase.
Step-by-step explanation:
(a) In a galvanic cell, the anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction occurs. In this battery, the copper electrode is undergoing oxidation, so it is the anode. The lead electrode is undergoing reduction, so it is the cathode.
(b) The half-reactions at the anode and cathode are:
At the anode (copper electrode): Cu(s) -> Cu²+(aq) + 2e-
At the cathode (lead electrode): Pb²+(aq) + 2e- -> Pb(s)
To find the overall reaction, we need to balance the two half-reactions:
2Cu(s) + Pb²+(aq) -> 2Cu²+(aq) + Pb(s)
(c) The standard cell potential for this battery can be determined by looking up the reduction potentials of the half-reactions and subtracting the reduction potential of the anode from the reduction potential of the cathode. The standard reduction potential for Cu²+(aq) + 2e- -> Cu(s) is +0.34 V, and the standard reduction potential for Pb²+(aq) + 2e- -> Pb(s) is -0.13 V. Subtracting the anode potential from the cathode potential gives us:
E°cell = E°cathode - E°anode
E°cell = (-0.13 V) - (+0.34 V) = -0.47 V
(d) As the battery operates, the concentration of Cu²+ ions in the copper half-cell will decrease while the concentration of Pb²+ ions in the lead half-cell will increase. This is because Cu²+ ions are being reduced to Cu(s) at the cathode, while Pb(s) is being oxidized to Pb²+ ions at the anode.