If an aqueous solution containing both potassium ethanoate and potassium propionate is subjected to Kolbe's electrolysis, nine products from the list below would be formed.
CH3CH3(I) CH3CH2CH3(II)
CH3CH2CH2CH3(III) CH3COOC2H5(IV)
CH3CH2COOC2H5(V) CH3COOCH3(VI)
CH3CH2COOCH3(VII)
CH2=CH2(IX) CH3CH=CHCH3(X)