Faraday's Calculation with 18 Volts: Why the Result is Inflated - Explanation

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If 18 volts are used in this experiment instead of 9.0 V, the value of the Faraday calculated is too large. Explain.

I don't get this question?
 
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If 18 volts are used in this experiment instead of 9.0 V, the value of the Faraday calculated is too large. Explain



well it Faraday's equal to 1 C = J/V which we have 1V = J/C that's why its too large. Is it right?