In the activity, click on the e∘cell and keq quantities to observe how they are related. use this relation to calculate keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. the two half-reactions that occur in the cell are cu2+(aq)+2e−→cu(s) and fe(s)→fe2+(aq)+2e− the net reaction is cu2+(aq)+fe(s)→cu(s)+fe2+(aq) use the given standard reduction potentials in your calculation as appropriate.

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Reduction at cathode :

Reduction potential of   to Cu=

Oxidation at anode:

Reduction potential of   to Fe=

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

The standard cell potential of the reaction is 0.78 Volts.

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 :

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .