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

am i too late?

explanation: