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I'm doing an experiment at school on the rate of reaction in galvanic cell so I wanted to question how the length of wires and the addition of different resistors could affect the rate of reaction. I'm a little confused with the fact that if I increase the resistance, according to the equation $$V=IR$$ current will increase. Theoretically does this mean that no matter how much resistance I provide, the voltage will be almost constant? Also, how does this explain the rate of reaction, since current isn't proportional to voltage because resistance is not consistent?

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closed as unclear what you're asking by Mithoron, aventurin, airhuff, Tyberius, Todd Minehardt Apr 19 '18 at 23:58

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  • $\begingroup$ Do you mean the current will decrease if the resistance increases? @Jess $\endgroup$ – Tyberius Apr 19 '18 at 18:34
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Theoretically does this mean that no matter how much resistance I provide, the voltage will be almost constant?

Mostly, yes, though the current will decrease as the resistance increase. The voltage of an ideal cell at the terminals is determined by the chemical species in the cell and their concentrations, according to the Nernst equation. The current that flows in response to this voltage is determined by the load (ie. resistor), up until the point that the cell becomes rate limited. The current directly describes the reaction rate because each mole of reaction results in 1 Coulomb of charge (or 2, 3, n, depending on the balanced equation). So if you have n amps of current, that's n coulombs/sec and 1 mol/sec.

In a real battery, there is some (small) resistance in the parts, such as the electrode. This means there will be a V=IR drop between the surface of the electrode, and the place you connect the volt-meter. So the measured voltage at high currents will be slightly lower.

At high rates, there comes a point where the battery cannot react fast enough to supply the current. One example is when the interface between the electrolyte and the electrode consumes ions faster than they can diffuse from the bulk. When this happens, the surface becomes depleted of ions and their effective concentration goes down, so the potential changes. Ultimately it's the concentration at the interface that matters.

Also, many (most) cells do not have a constant voltage over their entire charge states. If the concentrations (activities) change at all as the reaction happens, the potential will change as the cell discharges.

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