Using H2SO4 0.1M the current starts at about -2 mA and goes to about 3 mA..
And the perfil looks like a straight line in a 45º of inclination from x axis.
If the C/O ratio in graphite oxide is low (~ 2), the material is non-conductive which could explain why you get a CV typical for ohmic resistive materials.
If GO is non-conductive, why is the current value so high?? Like about 3000 uA at 1.3 V??
If the electrode's characteristic is mainly influenced by a high resistance, Ohm's law (E = R * I) applies for the electrode. As the resistance (R) is constant, you get a linear increase in current (I) with rising potential (E). The slope of the linear line is the resistance. Therefore, the observed current is only dependent from the resistance of the electrode. On the other hand, i would not use this kind of technique to determine the resistance of the material as the errors by using this kind of setup might be very high.
Here is the cyclic voltammogram of H2SO4 0.1M...
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