In the following reactions first identify the substance that is oxidized and the one that is reduced. then identify the oxidizing and reducing agents.
a. 2li(s)+ f2(g) --> 2lif(s)
b. fe(s) + cuso4(aq) --> feso4(aq)+ cu(s)

So the Li is oxidized and the F is reduced.
when you become negative you gain electrons=reduction

The atom that causes the reduction is the reducing agent, so since the Li caused the F to be reduced it is the reducing agent.And vice versa. The F is the oxidizing agent.

B) Fe is oxidized, but the element that is reduced is Cu.The charge must change for it to be oxidized or reduced.The SO4 is -2 the whole time.Fe is the reducing agent and Cu is the oxidizing agent.

Whatever is oxidized is the reducing agent, and whatever is reduced is the oxidizing agent.

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :