Find Ecell for an electrochemical cell based on the following reaction with [MnO4−]=1.20M, [H+]=1.50M, and [Ag+]=0.0100M. E∘cell for the reaction is +0.88V. MnO4−(aq)+4H+(aq)+3Ag(s)→MnO2(s)+2H 2O(l)+3Ag+(aq)

1.01 V

Explanation:

From Nernst equation;

Ecell= E°cell- 0.0592/n log Q

Where;

Ecell= observed emf of the cell

E°cell= standard emf of the cell

n= number of moles of electrons transferred

Q= reaction quotient

Q= [Ag^+]^3/[MnO4^-] [H^+]^4

Q= [0.01]^3/[1.20] [1.50]^4

Q= 1.65×10^-7

Ecell= 0.88 - 0.0592/3 log 1.65×10^-7

Ecell= 0.88 - [0.0197×(-6.78)]

Ecell= 0.88 + 0.134

Ecell= 1.01 V

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 :