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Reduce all coefficients to the lowest possible integers.
Include H2O(l) and H+(aq) appropriately to complete the equation.

1.Cr2O72−(aq) + H2C2O4(aq) → Cr3+(aq) + CO2(g)

2. CrI3(s) + Cl2(g) → CrO42−(aq) + IO4(aq) + Cl(aq)

Mar 20th, 2015

first equation

Cr2O72- + 14 H+ + 6e- = 2 Cr3+ + 7 H2O

H2C2O4 = 2 CO2 + 2 H+ + 2e- ) x 3

Cr2O7 ^2-(aq) + 3 H2C2O4(aq) + 8 H+(aq) = 2 Cr^(aq)+ + 6 CO2 (g)+ 7 H2O(l)

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second equation. 27 Cl2(g) + 2CrI3(aq) + 64 OH-(aq) --> 6 IO4-(aq) + 2 CrO4=(aq) + 54 Cl-(aq) + 32 H2O(l)
explanation
You should separate the equation into 2 half-equations:

CrI3 --> CrO4[-2] + IO4[-1]

Cl2 --> Cl[-1]

first, balance the more 'exotic' atoms such as Cl, I (anything besides H, O)
CrI3 --> CrO4[2-] + 3IO4[-]

Cl2 --> 2Cl[-]

balance O atoms by adding H2O on the other side.
CrI3 + 16H2O --> CrO4[2-] + 3IO4[-]
Cl2 --> 2Cl[-]

balance H atoms by adding H+ on the other side.
CrI3 + 16H2O --> CrO4[2-] + 3IO4[-] + 32H[+]
Cl2 --> 2Cl[-]

Balance charges by adding e- (electrons)
CrI3 + 16H2O --> CrO4[2-] + 3IO4[-] + 32H[+] + 27e-
Cl2 + 2e- --> 2Cl[-]

Add the same number of OH- as H+
Note that OH- reacts with H+ to form water.
CrI3 + 16H2O + 32 OH[-]--> CrO4[2-] + 3IO4[-] + 32H2O + 27e-

Cancel off all the h2o that appears on both sides.
CrI3 + 32OH[-] --> CrO4[2-] + 3IO4[-] + 16H2O + 27e-

Put the 2 half-equations together such that the number of e- on one side is equal to the other.
After that cancel all the e-.
CrI3 + 32OH[-] --> CrO4[2-] + 3IO4[-] + 16H2O + 27e-
Cl2 + 2e- --> 2Cl[-]
Combine -->
2CrI3 + 64OH[-] + 27Cl2--> 2CrO4[2-] + 6IO4[-] + 32H2O + 54Cl-

This is the final equation.
Note that this is the ionic equation for the reaction between chromium(III)iodide and chlorine gas to form chromate, periodate and chloride ions in basic medium. This is a redox reaction with Iodide being oxidised and chlorine being reduced. The equation for the same reaction occuring in acidic medium can be found by combining the 2 half-equations before the addition of OH- ions.

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There are two oxidation reactions occurring:
1. Cr+++ + 8 OH- ---------> CrO4= + 4 H2O + 3 e- and
2. I- + 8 OH- ----------> IO4_ + 4 H2O + 8 e-
one reduction reaction:
3. 2 e- + Cl2 ---------> 2 Cl-
To get an even no. of electrons in 1, take 1 x 2
4. 2 Cr+++(aq) + 16 OH-(aq) ----> 2 CrO4=(aq) + 8 H2O + 6 e-
for 2 Ce+++, there has to be 6 I-, multiply 2 by 6.
5 6 I-(aq) + 48 OH-(aq) + 27 Cl2(g) ---------> 6 IO4-(aq) + 24 H2O + 48 e-
Total electrons lost = 48 + 6 = 54 e-
6. Cl needed for 54 e- is 27 Cl2: 54 e- + 27 Cl2(g) -------> 54 Cl-
add 4,5 & 6
27 Cl2(g) + 2CrI3(aq) + 64 OH-(aq) --> 6 IO4-(aq) + 2 CrO4=(aq) + 54 Cl-(aq) + 32 H2O(l)

Mar 20th, 2015

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Mar 20th, 2015
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Mar 20th, 2015
Oct 19th, 2017
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