Ion-selective electrode: high input impedance to measure the voltage

Question

I read that an ion-selective electrode has an high impedance (\$100 \, M\Omega\$) and that it's necessary a high input impedance to measure the electrode voltage.

I'd like to understand the above sentence with a circuit. I tried to draw a circuit with a voltage source and two resistances (input resistance and load resistance):

for the Kirchhoff law:

$$E=(R_i+R_L) \, i$$

At this point it's difficult to me to understand why I need a high Ri. Could you suggest me the right way please?

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EDIT

I am studying a biosensor (https://ieeexplore.ieee.org/document/7576627/) and at page 947 there is the sentence: Ion-selective electrodes (ISEs) separate specific ions with a semi-permeable membrane between two electrodes, thereby creating a potential difference (∼0.1–100 mV) proportional to the amount of that ion concentration in the solution. However, due to the nature of these sensors, their impedance is very high, roughly on the order of 100 MΩ, necessitating high resolution sampling of the electrode voltage with a high input impedance.

Thank you in advance.

Answer 1

In your setup, the assumption is that both E and Ri are unknown. You are trying to measure E with a voltmeter having an input resistance of RL. The voltage seen by the voltmeter, call it Ev, will be the result of the voltage divider formed by Ri and RL, namely Ev = E*RL/(RL+Ri). Since Ri is not known, E cannot be accurately determined unless RL is much greater than Ri. In that case, the equation reduces to Ev = E. That is why your reference calls for a high input impedance voltmeter.