A cell is set up with copper and lead electrodes in contact with CuSO₄(aq) and Pb(NO₃) ₂(aq) ,respectively,at 25°C.The standard reduction potentials are: Pb²⁺ + 2e^- $\to$ Pb $\varepsilon$ ^$\circ$ = -0.13 V
Cu²⁺ + 2e^- $\to$ Cu $\varepsilon$ ^$\circ$ = +0.34 V
If the Pb²⁺ and Cu²⁺ are each 1.0 M,the potential of the cell,in volts,is:

Question

A cell is set up with copper and lead electrodes in contact with CuSO₄(aq) and Pb(NO₃) ₂(aq) ,respectively,at 25°C.The standard reduction potentials are: Pb²⁺ + 2e^-

\to

Pb

\varepsilon

^$\circ$ = -0.13 V
Cu²⁺ + 2e^-

\to

Cu

\varepsilon

^$\circ$ = +0.34 V
If the Pb²⁺ and Cu²⁺ are each 1.0 M,the potential of the cell,in volts,is:

Quizplus · Accepted Answer

The potential of the cell is calculated using the standard reduction potentials of the two half-reactions. The formula to calculate the cell potential (E°cell) is E°cell = E°cathode - E°anode. Here, the copper (Cu) electrode acts as the cathode with a standard reduction potential of +0.34 V, and the lead (Pb) electrode acts as the anode with a standard reduction potential of -0.13 V. Therefore, E°cell = +0.34 V - (-0.13 V) = +0.47 V.

A Cell Is Set Up with Copper and Lead Electrodes →\to→

A Cell Is Set Up with Copper and Lead Electrodes $\to$