How to write this formula?

Question

A solution with I- and a sour solution with Cr2O72- is mixed. Cr3+ and I is the result of this. What is the formula for this reaction?

I basically don't know how to balance it.. The unbalanced thing I presume is Cr2O72- + Cr3+ --> I + Cr3+. Then, I'm lost!

Sorry, got the numbers mixed up.. I of course didnt mean to write Cr3+ on both sides.. I- on the first!

Answer 1

Using half-equations:-
2I- = I2 + 2e-
Cr2O7 2- + 14H+ + 6e- = 2Cr3+ + 7H2O

These are standard half-equations - memorise them. !!!!

NB there are 2 electrons (RHS) in the upper equation, and
there are 6 electrons (LHS) in the lower equation.
The number of electrons is equalised by multiplying the upper equation by 3. So we have:-
6I- = 3I2 + 6e-
Cr2O7 2- + 14H+ + 6e- = 2Cr3+ + 7H2O

Next remove one of the lots of six electrons across the 'equals' sign not forgetting to change the sign in front of the electrons (Just like maths).
Hence
6I- - 6e- = 3I2
Cr2O7 2- + 14H+ + 6e- = 2Cr3+ + 7H2O
Adding these two half- equations - the electrons are eliminated.

Cr2O7 2- + 14H+ + 6I- = 3I2 + 2Cr3+ + 7H2O
NB
1. The charges balance
2 - + 14+ + 6- = 2 x 3+ = 6+
2. The atoms balance
2 x Cr = Cr x 2
7 x O = O x 7
14 x H = 7 x H2
6 x I = 3 x I2

Answer 2

On the right side the chrome is still oxydized but in the form Cr2O3. The four more oxygen atoms on the left combine with the H+ available in the acid (not sour in English) solution. Therefore you should balance
H+ + Cr2O7 2- + I- -> Cr2O3 + I2 + H2O

Answer 3

I think it's imposibble there's a compound of iodine exists in I, because it's diatomic substance, it should appear in I2.

So, the way to balance it is..

6 e- + Cr2O72- + 14 H+ --> 2 Cr3+ + 7 H2O | x1
2 I- --> I2 + 2 e- | X3
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Cr2O72- +14 H+ + 6 I- --> 2Cr3+ + 3 I2 + 7 H2O

This reaction needs acid and produce water ( H2O)

Answer 4

you have to take in account 'I' in the reactants' side also and try balancing it.