The equilibrium constant for the reaction 2MnO4- + 6H+ + 5H2C2O4gives 2Mn+2 + 8H2O + 10CO2is approximately.given that E?CO2/C2O4--=-0.49 V and E?MnO4-/Mn+2 =1.51 V
Dear student
Please find below the solution of your asked query.
EOcell=Ecathode-Eanode=1.51+0.49=2 volts
at equilibrium Ecell =0
so E0Cell = -0.059/n log kc
2= -0.059/5log kc
10/-0.059 =log kc
kc =antilog 0.40= 2.51
Hope this solution clears your concept regarding the topic.
Thanks