# Why does increasing surface area of the electrode decrease resistance?

Which increases immediately if the surface area of the silver electrode is increased?

1. overall cell voltage
2. rate of change of $\ce{[Ag+]}$
3. mass of $\ce{Cu}$ electrode
4. change in ratio of electrode masses; ∆$(\frac{\text{ mass of Cu}}{\text{ mass of Ag}})$

The correct answer was (2), but I'm not sure why the reaction rate would go up if the surface area of the silver electrode increased. The rate of reaction is proportional to the current flowing through, which by Ohm's Law is $\frac{V}{R}$. This means that the resistance must have decreased if current increases. But resistance of the electrodes is only related to the length and cross sectional area. The problem does not specifically state any change in length or cross sectional area. It only says surface area, so what am I missing here?

• For a galvanic cell (electrolytic cell with significant current) $\mathrm{R} \propto \dfrac{1}{\text{Area(electrode)}}$ because of the overpotential toruhara.page.tl/… – MaxW Mar 22 '16 at 0:16
• There is a greater point of contact with the larger electrodes allowing the current to flow at a faster rate. – Technetium Mar 22 '16 at 1:58
• Well actually in the circuit that you show, a high impedance voltmeter essentially allows no current to flow and you are just measuring the "theoretical" EMF of the cell. – MaxW Mar 22 '16 at 5:15