On the sheet Mr Beaumont gave us, Q13 is
Cr2O7 2- + 14 H+ + 6 e- ---> 2Cr3+ + 7H2O
Fe2+ ----> Fe3+ + e-
60 cm3 of acidified dichromate (VI) solution with a concentration of 0.05 mol/dm3 was titrated against a 0.6 mole/dm3 Fe 2+ solution. What volume of Fe 2+ solution would be required to reach the end point of this titration?
I don't understand how the equations above help you to work out what happens in the titration except for the fact you can see what the ions of each substance are...
Thanks.
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The bit of the equations that is important is the number of electrons. When you combine the 2 half-equations, you must be able to cancel out the electrons so they don't appear in the overall equation.
This means that there must be the same number of electrons in each half-equation. What do you think you need to do if they are different?
Once you've made sure that there are the same number of electrons in each half-equation, the equations should give you the ratio in which the 2 reactants react with each other.
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