When the reaction Al (s) + Cr2O72- (aq) → 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
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!
To balance the given reaction in acid, let's break it down into half-reactions:
1. Reduction half-reaction:
Cr2O7^2- (aq) → Cr^2+ (aq)
2. Oxidation half-reaction:
Al (s) → Al^3+ (aq)
Now, let's balance the half-reactions:
1. Reduction half-reaction:
Cr2O7^2- (aq) + 14 H+ (aq) + 6 e- → 2 Cr^2+ (aq) + 7 H2O (l)
2. Oxidation half-reaction:
2 Al (s) → 2 Al^3+ (aq) + 6 e-
To balance the electrons in the overall reaction, we need to multiply the oxidation half-reaction by 3:
6 Al (s) → 6 Al^3+ (aq) + 18 e-
Adding the two half-reactions together:
Cr2O7^2- (aq) + 14 H+ (aq) + 6 e- + 6 Al (s) → 2 Cr^2+ (aq) + 7 H2O (l) + 6 Al^3+ (aq) + 18 e-
Simplifying:
Cr2O7^2- (aq) + 14 H+ (aq) + 6 Al (s) → 2 Cr^2+ (aq) + 7 H2O (l) + 6 Al^3+ (aq)
The balanced equation shows that 14 protons (H+) are consumed. Therefore, the correct answer is option C: 8 protons (H+) are consumed.
To balance the given reaction, Al (s) + Cr2O72- (aq) → Cr2+ (aq) + Al3+ (aq), you need to ensure that the number of atoms of each element is the same on both sides of the equation. In addition, since the reaction is in acid, protons (H+) are involved.
Here's a step-by-step explanation:
1. Start by determining the oxidation states of each element in the reactants and products. In this case, chromium (Cr) goes from a +6 oxidation state in Cr2O72- to a +3 oxidation state in Cr2+. Aluminum (Al) goes from a 0 oxidation state in Al (s) to a +3 oxidation state in Al3+.
2. Determine the number of each element in the reactants and products. There are two chromium atoms in Cr2O72- and Cr2+, and one aluminum atom in Al (s) and Al3+.
3. Begin balancing the equation by balancing the atoms of each element, excluding hydrogen and oxygen. In this case, both sides of the equation have two chromium atoms and one aluminum atom, so they are already balanced.
4. Next, balance the oxygen atoms by adding water (H2O) molecules to the side that needs additional oxygen. In this case, there are seven oxygen atoms in Cr2O72-, so add seven water molecules to the reactant side.
Cr2O72- + 7H2O → Cr2+ + Al3+
5. Now, balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs additional hydrogen. In this case, there are fourteen hydrogen atoms added from the seven water molecules, so add fourteen hydrogen ions to the reactant side.
Cr2O72- + 7H2O + 14H+ → Cr2+ + Al3+
6. Finally, check and adjust the charges to ensure they are balanced. In this case, the charges are balanced with a total charge of +12 on both sides.
The equation is now balanced as:
2Cr2O72- + 7H2O + 14H+ → 2Cr2+ + 6H2O + Al3+
From this balanced equation, we can see that 14 protons (H+) are consumed. Therefore, the correct answer is B. 3 protons (H+) are consumed.