2.2 mL is how many mililiters of active ingredient
With the information given most likely in order to find the partial pressure of the gas produced you have to subtract the total air pressure in the collection flask by the atmospheric pressure since you assume that the flask started with atmospheric pressure when it was sealed and then the gas was added as the reaction took place increasing the pressure.
1.44atm-0.95 atm=0.49atm
Answer:
E°= E°cathode- E° anode= 0.271-0.330= -0.59V
Explanation:
NB: the stoichiometry does not affect E°values,
And the more positive the E° values , the greater it's tendency to become spontaneous and hence irreversible, and the more negative the E° values the more likely to become less spontaneous and reversible, hence the above reaction is reversible
The ion in the cathode that gains electrons
