A voltaic cell consists of Cu(2+)/Cu(2+) and Cd(2+)/Cd half-cells.


Screenshot 2024 01 28 131939

Screenshot 2024 01 28 131939

Screenshot 2024 01 28 131950

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​+2H2​O

 

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:

  1. Therefore, the concentration of Cu2+ when the voltage drops to 0.719V is approximately 0.0707 M.
  2. 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.

 

 

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