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It means that it doesn't matter how many times the reaction occurs, E will remain the same as long as it is at standard conditions. Unlike in something like enthalpy or Gibb's free energy where the standard formation value gets multiplied by the moles of the reaction, E will remain the same for one mole or multiple moles of reaction. Just like how when you have four moles of water or one mole of water, the density of water will be the same for both.
Yes, so the importance of this property comes in when you are balancing your redox half reactions. In redox, you want to make sure that the amount of electrons lost is equal to the amount of electrons that is gained. Just like other problems from the last section, you use molar coefficients to balance the redox reactions. However, what is different is that when you add a molar coefficient, unlike hess's law problems, the value of E is not multiplied by the molar coefficient because it is intensive. Instead, even with a greater number of products and reactants, the value of E is the same
The key takeaway is that you dont change the E standard cell potential value whenever you balance equations. We're used to multiplying by a factor or dividing by a factor when balancing more than 1 equation (ex. enthalpy change) but for Ecell, leave the value as is and just add or subtract values.
E° is intensive basically means that it doesn't rely on how many times the reaction occurs. The voltage difference will always remain the same, which is why in the example that Dr. Lavelle did on that day, he did not multiply E° by the coefficient to balance the equation.
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