Represent the galvanic cell in which the reaction takes place.
$Z n (s) + {C u}^{2 +} (a q) \to {Z n}^{2 +} (a q) + C u (s)$

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Solution

Measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials.
This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through.
If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when $Z n^{2 +}$ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the $C u^{2 +}$ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:
$Z n_{(S)} + C u_{(a q)}^{2 +} ⟶ Z n_{(a q)}^{2 +} + C u_{(s)}$

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