Asked by Anonymous
                When the reaction Al (s) + Cr2O72- (aq) β Cr2+ (aq) + Al3+ (aq) is correctly balanced in acid, Which one is it? Please explain.
A. 7 protons (H+) are consumed
B. 3 protons (H+) are consumed
C. 8 protons (H+) are consumed
D. 42 protons (H+) are consumed
E. 12 protons (H+) are consumed
            
        A. 7 protons (H+) are consumed
B. 3 protons (H+) are consumed
C. 8 protons (H+) are consumed
D. 42 protons (H+) are consumed
E. 12 protons (H+) are consumed
Answers
                    Answered by
            Cole
            
    It is 42 H+ consumed. Here is the step by step:
1) break into Ox/Red reactions.
Ox: Al(s) -> Al3+(aq)
Red: C2O7 2- (aq) -> Cr2+
Balance Redox Reaction (Ox is balanced other than charge)
1st Balance Oxygen by adding H2O
C2O7 2- -> Cr2+ +7H2O
Next, balance Hydrogens by adding H+
(in acidic solution so no need to neutralize with OH-)
C2O7 2- +14H+ -> Cr2+ +7H2O
Finally, Balance charges in Red and Ox
Red: C2O7 2- +14H+ +8e- -> Cr2+ + 7H2O
Ox: Al -> Al3+ +3e-
Balance electrons by multiplying each reaction so that electrons cancel.
Red (x3) = 3C2O7 2- +42H+ +24e- -> 3Cr2+ +7H2O
Ox (x8) = 8Al -> 8Al3+ + 24e-
Finally, add equations to get the total reaction (and so electrons cancel).
3C2O7 2- + 8Al + 42H+ -> 8Al3+ + 3Cr2+ + 7H2O
Therefore, 42 H+ are consumed in the reaction.
Sorry for the bad formatting, hope this helps!
    
1) break into Ox/Red reactions.
Ox: Al(s) -> Al3+(aq)
Red: C2O7 2- (aq) -> Cr2+
Balance Redox Reaction (Ox is balanced other than charge)
1st Balance Oxygen by adding H2O
C2O7 2- -> Cr2+ +7H2O
Next, balance Hydrogens by adding H+
(in acidic solution so no need to neutralize with OH-)
C2O7 2- +14H+ -> Cr2+ +7H2O
Finally, Balance charges in Red and Ox
Red: C2O7 2- +14H+ +8e- -> Cr2+ + 7H2O
Ox: Al -> Al3+ +3e-
Balance electrons by multiplying each reaction so that electrons cancel.
Red (x3) = 3C2O7 2- +42H+ +24e- -> 3Cr2+ +7H2O
Ox (x8) = 8Al -> 8Al3+ + 24e-
Finally, add equations to get the total reaction (and so electrons cancel).
3C2O7 2- + 8Al + 42H+ -> 8Al3+ + 3Cr2+ + 7H2O
Therefore, 42 H+ are consumed in the reaction.
Sorry for the bad formatting, hope this helps!
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