Oxidation state of so4

Sulfate is a covalent ion. So it's more realistic to use formal charges, not oxidation states, to describe its atoms. It's OK to give sulfate an overall oxidation state, but it doesn't make sense to do so for sulfur and oxygen in sulfate. OXIDATION STATE APPROACH: IDEAL IONIC BONDS Oxidation states are hypothetical charges for a compound that has ideal ionic bonds, or would form ideal ionic bonds, i.e. complete transfer of the electron pair to only one atom in the bond. You wrote #"SO"_4^(2-)#, which is the sulfate polyatomic ion. You wrote that its total charge was #-2#. Since sulfate ( #"SO"_4^(2-)#) is already ionized, its charge is its oxidation state. So it's kind of pointless to find the "oxidation state" for #"SO"_4^(2-)# because it's already given. The oxidation states for sulfur and oxygen atom are determined from: • electronegativities • total charge #"SO"_4^(2-)# has a total charge of #-2#, and oxygen is more electronegative than sulfur. So, oxygen polarizes the electron more negative oxidation state here. Since oxygen is on column 16, its expected oxidation state is #color(red)(-2)#. Therefore: #-2xx4 + x = -2# #=># #x# gives the oxidation state of sulfur, which is #color(red)(+6)#. #=> color(highlight)(stackrel(color(red)(+6))("S")stackrel(color(red)(-2))("O"_4^(2-)))# But keep in mind that this is unrealistic. If you ever do this for redox balancing, it's only an accounting scheme, and nothing more. FORMAL CHARGE APPROACH: IDEAL COVALENT BONDS #"SO"_4^(2-)# i...

$\ce$ per hour. Assume efficiency 50%. I was solving this question on Faraday's Laws Of Electrolysis when I stumbled across a conceptual flaw of mine. I realized that the sulfate anion in the first reaction is at a +6 Oxidation State and so is Marshall's acid, as far as what I know of Redox Reactions and their balancing we look at the number of electrons exchanged and thus formulate the half-cell reaction. However, this logic of mine failed in the above question as the oxidation state of the central atom is unchanged, which makes me wonder how to calculate the valency-factor/N-factor and correspondingly the equivalent weight. I know this is a conceptual shortcoming of mine and that the aforementioned logic is very 'methodical' per se, which is why it fails. If someone could please point out where I am going wrong it'd be highly appreciated. Forget about the equivalent weights. Work with moles, and only with moles. Look how it goes. $102$ g $\ce$$ This value is obtained if the yield is $100$%. As the yield is $50$%, the intensity must be twice the previous value. This is $320$ A. In the event that this question is actually more than just a theoretical exercise, the general electrolysis half-reactions, as cited by Maurice in my opinion, may not actually be in accord with this particular's complex reaction system per a review of the possible underlying mechanics. More precisely, here is a suggested overview of reaction mechanics that support my comment: $\ce$ in an experiment...

\( \newcommand\) • • • Acid Equilibria \[\ce\) is a very strong acid. Because it is such a weak base, sulfate ion undergoes negligible hydrolysis in aqueous solution. Solubility Most sulfates, including those of \(\ce\), even if it is acidic, is a reliable test for sulfate. Other insoluble sulfates are those of calcium, strontium, and mercury(I).

\( \newcommand\) • Oxidation-Reduction or "redox" reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers. The Half Equation Method is used to balance these reactions. In a redox reaction, one or more element becomes oxidized, and one or more element becomes reduced. Oxidation is the loss of electrons whereas reduction is the gain of electrons. An easy way to remember this is to think of the charges: an element's charge is reduced if it gains electrons (an acronym to remember the difference is LEO = Lose Electron Oxidation & GER = Gain Electron Reduction). Redox reactions usually occur in one of two environments: acidic or basic. In order to balance redox equations, understanding Some points to remember when balancing redox reactions: • The equation is separated into two half-equations, one for oxidation, and one for reduction. • The equation is balanced by adjusting coefficients and adding H 2O, H +, and e - in this order: • Balance the atoms in the equation, apart from O and H. • To balance the Oxygen atoms, add the appropriate number of water (H 2O) molecules to the other side. • To balance the Hydrogen atoms (including those added in step 2), add H + ions. • Add up the charges on each side. They must be made equal by adding enough electrons (e -) to the more positive side. • The e - on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made ...