Many different ways have been proposed to make batteries. One cell is set up with copper and lead electrodes in contact with CuSO₄(aq) and Pb(NO₃)₂ (aq), respectively. If the Pb²⁺ and Cu²⁺ concentrations are each 1.0 M, what is the overall cell potential $\begin{array}{ll} \mathrm{Pb}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb} & \mathrm{E}^{\circ}=-0.22 \mathrm{~V} \ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} & \mathrm{E}^{\circ}=+0.34 \mathrm{~V} \end{array}$

Question

Quizplus · Accepted Answer

The cell potential can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is Faraday's constant, and Q is the reaction quotient.

The standard cell potential can be looked up in a table (or calculated using standard reduction potentials of the half-reactions involved):

E°cell = E°(Cu2+/Cu) - E°(Pb2+/Pb) = 0.34 V - (-0.13 V) = 0.47 V

The reaction quotient can be calculated using the concentrations given:

Q = [Cu2+]/[Pb2+] = 1.0/1.0 = 1

Plugging in the values:

Ecell = 0.47 V - (0.0257 V/K)(298 K)/(2)(96,485 C/mol)ln(1) 
= 0.47 V - 0 
= 0.47 V

Therefore, the overall cell potential is 0.47 V, which corresponds to answer choice C.

Many Different Ways Have Been Proposed to Make Batteries A) 016 V B) 0

Many Different Ways Have Been Proposed to Make Batteries A) 016 V
B) 0