Question

plzzz give me question 13 without any spam


If u spam then u can understand ​

Answer 1

13 a) in KMnO4 usual oxid. satate of O is -2 and for K it is +1

hence oxidation state of Mn be x=  x-2(4)+1=0

x=+7

hence oxid of Mn=+7

similarly in H2S2O7

usual oxid state of H=+1,O=-2

hence 2x-2(7)+1(2)=0

2x=12

x=+6

oxid state of s=+6

in H3PO3

x+1(3)-2(3)=0

x=+3

oxid state of P=+3

part (b)

value of emf can be calculated=emf at reduction-emf at oxidation

emf=-0.25-(-0.76)

emf of cell=0.51 volts