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Chemistry: Post your doubts here!

A

ashcull14

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In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0cm3 of 0.10moldm–3 aqueous sodium sulphite. The half-equation for oxidation of sulphite ion is shown below.
SO3^2− (aq) + H2O(I) → SO4^2− (aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?
 
princess Anu

princess Anu

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ashcull14 said:
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If you find the ratio of salt to sulphite( simply by finding moles of salt and sulphite according to the info given) it comes out to be 2:1
& the equation shows 1 mole of sulphite looses 2 electron, therefore 2 mol of metallic salt gain those 2 electrons or 1 mol of metallic salt gains 1 mol of electron
therefore new O.N. of metal = +3-1=+2!
 
princess Anu

princess Anu

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lara dalal said:
Hello. How do i do this? The answer is D
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for the mercury level to rise at right end, the pressure being exerted from gases in bulb R should increase!
Now in option 1,if temp is increased, forward reaction is favoured, and will lead to the formation of 2 moles of the gas which will inturn exert greater pressure. However in option 2, there are equal no of moles on both sides of the equilibrium so even if as temp is increased backward reaction may be favoured but it'll not cause an INCREASE in pressure.
 
A

ashcull14

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princess Anu said:
If you find the ratio of salt to sulphite( simply by finding moles of salt and sulphite according to the info given) it comes out to be 2:1
& the equation shows 1 mole of sulphite looses 2 electron, therefore 2 mol of metallic salt gain those 2 electrons or 1 mol of metallic salt gains 1 mol of electron
therefore new O.N. of metal = +3-1=+2!
Click to expand...
thnk u SO much that was very helpful indeed :) <3
 
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lara dalal

lara dalal

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ashcull14 said:
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H3PO4 is generally not an oxidising agent thats why options A and B are wrong. For option C, During the reaction of NaCl and H2SO4, HCl (the acidic gas) is formed, if H2SO4 was a stronger oxidising agent it would go on and form the yellow/green gas of Cl2. but it didnot hence it is NOT a stronger oxidising agent. If H2SO4 was a stronger oxidising agent it would have Completely oxidised the Cl- to Cl2. For D NaI Reacts with H2SO4 to form HI, the HI is further oxidised by the H2SO4 to form the purple vapour of iodine; the I- got oxidised to I2 This shows that H2SO4 IS in. fact a stronger oxidising agent than iodine
 
lara dalal

lara dalal

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Awesome12 said:
Why can't it be C?

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How could it be C. The OH will combine with the H beside it to get removed as H20 and leave a double bond behind. In C the double bond formed is 2 carbons away from the position of OH. So the OH Could never reach the Hydrogen on that carbon 2 places away from it. In D the OH combines with the H on the Carbon right next to it ( they just showed a rotation in the figure in D to confuse you) I hope you understood what i mean
 
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