Final answer:
To increase the cell potential of the voltaic cell, option (a) increase the concentration of Pb²⁺ should be chosen. This is because increasing the concentration of reactants where reduction occurs or decreasing the concentration where oxidation occurs will make the cell potential more positive according to the Nernst equation.
Step-by-step explanation:
To make the cell potential more positive in a voltaic cell, you have to manipulate the concentrations of the ions involved according to the Nernst equation. In a concentration cell, like the one being considered with lead (Pb) and copper (Cu) species, increasing the concentration of the reactants in the half-cell where reduction occurs or decreasing the concentration of the reactants where oxidation occurs will increase the cell potential.
For the specific cell mentioned, increasing the concentration of Pb²⁺ would push the reaction towards more lead being reduced, which increases the cell potential, according to the Nernst equation. Thus, the correct option to make the cell potential more positive is (a) increase [Pb²⁺]. Mentioning this correct option in the final answer, among the given choices, it's clear that option (a) would result in an increased cell potential for the voltaic cell in question.