### Step 1 of 2

Given data: Part A. Given the standard cell potentials (E∘) for the two half-cells:

E∘Cu=0.340V

E∘Cd=−0.400V

The initial concentration of Cu2+=1.250M and the initial concentration of Cd2+=0.250M

volume=0.500 l

Part B. Given is the reaction

PbO2+Pb+2H2SO4→2PbSO4+2H2O

Find: Part A. find the concentration of Cu2+

Part B. find the correct option

**Explanation:**

To determine the concentration of Cu2+ when the voltage has dropped to 0.719V, we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard cell potential E°cell, the concentration of reactants and products, and the number of electrons transferred in the balanced equation.

### Step 2 of 2

Part A.

To determine the concentration of Cu2+ when the voltage drops to 0.719V, we can use the Nernst equation:

Ecell=E°cell−(0.0592n)×log(Q)

Where:

Ecell = cell potential under non-standard conditions

E°cell = standard cell potential

n = number of electrons transferred

Q = reaction quotient

First, calculate the initial cell potential using the given standard reduction potentials:

E°cell = E°(cathode) – E°(anode)

E°cell = 0.340V−(−0.400V)=0.740V

Now, using the Nernst equation:

0.719V=0.740V−(0.05922)×log(Q)

Solving for log(Q):

−0.021V=(0.05922)log(Q)−0.021V=0.0296 log(Q)log(Q)=−0.7081Q=10−0.7081

Now, use the concentrations of Cu2+ and Cd2+ to find the concentration of Cu2+ when the voltage drops to 0.719V

Q=[Cu2+][Cd2+]10−0.7081=[Cu2+]0.250[Cu2+]=10−0.7081×0.250[Cu2+]=0.0707 M

Therefore, the concentration of Cu2+ when the voltage drops to 0.719V is approximately 0.0707 M.

Part B.

From the reaction equation:

PbO2+Pb+2H2SO4→2PbSO4+2H2O

During the use of the battery, the following processes occur:

- PbO2 is reduced to Pb at the cathode.
- PbSO4 is formed at the anode.

In the given reaction, during the use of the lead storage battery, lead (Pb) is oxidized at the anode to form lead ions (Pb2+), and lead dioxide (PbO2) is reduced at the cathode to form lead sulfate (PbSO4).

The correct statement is:

(Pb2+ is formed at the cathode during use of the battery.

**Explanation:**

- Therefore, the concentration of Cu2+ when the voltage drops to 0.719V is approximately 0.0707 M.
- Therefore, option b is the correct statement.

##### Final solution

Hence we can conclude,

Part A. The concentration of Cu2+ when the voltage drops to 0.719V is approximately 0.0707 M.

Part B. The option b is the correct statement.