The I in I2 is zero and it changes to +1 on the right.
The I in KIO3 is +5 and it goes to +1 on the right.
The easiest way to do this is to set up two half equations.
I2 ==> 2ICl
IO3^- ==> ICl
Balance those two, multiply by the appropriate numbers, add them, then cancel anything common to both sides.
on this problem, the e- balance, but when I start trying to balance the equation, I am not getting the same amount of I or Cl on each side. I can add H^+ or OH- or H2O, but I don't think I can add more elements. Please help me!
I would do it this way.
I2 ==> 2ICl
IO3^- ==> ICl
You CAN add ions if they are spectator ions (they don't change oxidation state) and you want a molecular equation instead of an ionic equation.
For I2 ==> 2ICl
I2 is zero on the left: +1 on the right(for each) or +2 for both. (That's the first mistake you made). Now we add electrons.
I2 ==> 2ICl + 2e
Now I would add the Cl^- like so.
I2 + 2Cl^- ==> 2ICl + 2e
Adding 2Cl^- is the only way to get them into the equation.
Charge on left is -2 (from 2Cl^-) and on the right is -2 (from 2e) and this half is balanced.
The other one is the following and you need to do it. Here is what you should get.
6H^+ + 4e + Cl^- + IO3^- ==> ICl + 3H2O
Following those steps I wrote in the previous problem will get these things done almost every time. I have found, through the years, one or two that I've had trouble with when I use this method.
1gm of AgNO3 is dissolved in 50ml of h2o. it is titrated with 50ml od sol. agi is precipitated is filtered of.the exces of ki is then titrated with 1/10M KIO3 in the presence of acidic medium until all I->>ICl. it reqires
HIO3 + FeI2 + HCl --> FeCl3 + ICl (in acidic solution) Show and label the two balanced half-reactions (oxidation and reduction). Add H2Os, H+s or OH-s as needed. I just wanted to know if the answer I got is correct 2FeI2 +6HCl
An anyhdrous metal chloride, X, contains 37.4% Cl. X is a reducing agent, 0.895 g requiring 23.60 cm^3 of 0.100M KIO3 for complete oxidation in 3- 9M HCl. X reacts with potassium chloride to form an adduct Y, containing 14.8% K.
Calculate the number of moles of KIO3 consumed by reaction (2.1). Remember, it is the limiting reagent. Using this information, calculate the number of moles of I- that must have been consumed by reaction. Unbalanced equation: IO
What mass of KIO3 is needed to convert the copper in 0.200g of CuSO4.H2O to Cu(IO3)2? my answer is: 0.828g Cu(IO3)2 x (2*214.003 MolarM KIO3/ 413.36 MolarM Cu(IO3) = 0.857g KIO3 could you please check if i got it right
If 32.3 grams of KCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass? 5.94% KCl 8.55% KCl 14.4% KCl 16.8% KCl Can someone explain this to me? I don't want the answer I just want to
If 6.01 g of K2SO3 dissolved in 930 mL of water reacts stoichiometrically according to the balanced equation in a reaction solution with a total volume of 990 mL, what is the molarity (M) of KCl produced? K2SO3(aq) + 2 HCl(aq)