Half-Peactions for the following oxidation-reduction reaction that occurs in acid solution. Determine the number of electrons that appears in the balanced half reactions. Cr20, (aq) + Cl (aq) → Cr* (aq) + Cl2 (9) Number of electrons =
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7) The reaction given is,
=> Cr2O72- (aq) + Cl- (aq) --------> Cr3+ (aq) + Cl2 (g)
In the above reaction, Cr is initially in 6+ oxidation state in Cr2O72- (because each O is in 2- oxidation state and there is 2- charge on the ion) and finally its in 3+ oxidation state as Cr3+
And Cl is in -1 oxidation state initially as Cl- and finally in 0 oxidation state in Cl2
Since the oxidation state of Cr is decreasing. Hence Cr is being reduced in the reaction.
And since the oxidation state of Cl is increasing in the reaction. Hence Cl is being oxidised in the reaction.
Hence the oxidation and reduction half reactions can be written as,
Oxidation half : Cl- (aq) --------> Cl2 (g) + e-
Reduction half : Cr2O72- (aq) + e- --------> Cr3+ (aq)
Balancing : Oxidation half : Since we have 2 Cl in RHS. Hence making it 2 in LHS also.
=> 2 Cl- (aq) --------> Cl2 (g) + e-
Now we have 2- charge in LHS. Hence making it 2- in RHS also.
=> 2 Cl- (aq) --------> Cl2 (g) + 2 e-
Since both charge and number of elements are equal in both side of the reaction.
Hence the above reaction is balanced oxidation half reaction.
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