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Posted in Chemistry

Balance this redox reaction occurring in acidic media: Cr2O7[2-](aq) + Cl[-](aq) Cr[3+] (aq) + Cl2(g)

Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules. 
Cr2O7[2-](aq) +2 Cl[-](aq) < – >2 Cr[3+] (aq) + Cl2(g)+7H2O 
Presently adjust that charges. 
you have – 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges 
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < – >2 Cr[3+] (aq) + Cl2(g)+7H2O 
take note of that the oxidation number of hydrogen in water is +1

Posted in Chemistry

Balance the following redox equation, using half-reactions. Assume that the reaction occurs in an aqueous solution. Cr2O72– + NO → Cr3+ + NO3–

First determine the formal oxidation numbers: 
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3 
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons: 
Oxidation half reaction: 
NO(aq) + 2 H2O(l) —> (NO3)-(aq) + 4 H+(aq) + 3 e- 
Reduction half reaction: 
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- —> 2 Cr3+(aq) + 7 H2O(l) 
Now balance the number of electrons on both sides and add them together: 
2 NO(aq) + 4 H2O(l) —> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- —> 2 Cr3+(aq) + 7 H2O(l) ————————————–… 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) —> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l) 
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly: 
-2 + 6 = -2 + 2(+3) +4 = +4 

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