Fe2+(aq) + Cr2O7 2- (aq) →Fe3+(aq) + Cr3+(aq)
Balance the equation by using oxidation and reduction half reactions. What is smallest possible integer coefficient of Cr3+ in the combined balanced equation?
Fe^+2 ==> Fe^+3 + e
Cr2O7^2- +6e + 14H^+ ==>2Cr^3+ + 7H2O
You finish. Multiply equation 1 by 6 and equation 2 by 1, then add the two.
To balance the given equation using oxidation and reduction half-reactions, let's break it down into two half-reactions: oxidation and reduction.
Oxidation half-reaction:
Cr2O7 2- → Cr3+
Reduction half-reaction:
Fe2+ → Fe3+
Let's start by balancing the oxidation half-reaction:
1. Determine the oxidation states of Cr and O in Cr2O7 2-.
Cr: Since O is generally assigned an oxidation state of -2, and there are seven O atoms, the total oxidation state contributed by O is -14. Therefore, the oxidation state of Cr can be determined by setting the sum equal to -2 (the overall charge of the ion):
2x + (-14) = -2
2x - 14 = -2
2x = 12
x = 6
So, Cr has an oxidation state of +6.
2. Since Cr has an oxidation state of +6 in Cr2O7 2- and +3 in Cr3+, it loses 3 electrons during the reaction. Rewrite the Cr2O7 2- oxidation half-reaction with the electrons included:
Cr2O7 2- → 2Cr3+ + 3e-
Now, let's balance the reduction half-reaction:
1. Determine the oxidation states of Fe.
Fe2+ has an oxidation state of +2, and Fe3+ has an oxidation state of +3.
2. Since Fe has an oxidation state of +2 in Fe2+ and +3 in Fe3+, it loses 1 electron during the reaction. Rewrite the Fe2+ reduction half-reaction with the electrons included:
Fe2+ → Fe3+ + 1e-
Now let's balance the number of electrons transferred in both half-reactions. We need to multiply each half-reaction by an appropriate factor to balance the electrons:
Multiplying the oxidation half-reaction by 2:
2Cr2O7 2- → 4Cr3+ + 6e-
Multiplying the reduction half-reaction by 3:
3Fe2+ → 3Fe3+ + 3e-
Now we can combine the half-reactions to form the overall balanced equation:
2Cr2O7 2- + 3Fe2+ → 4Cr3+ + 6Fe3+
From the balanced equation, we can see that the smallest possible integer coefficient of Cr3+ is 4.
To balance the given equation, we need to identify the oxidation and reduction half-reactions.
Let's start with the half-reaction for the oxidation of Fe2+ to Fe3+:
Fe2+ (aq) → Fe3+ (aq)
To balance the atoms, we need to add 1 electron on the left side:
Fe2+ (aq) + 1e- → Fe3+ (aq)
Next, let's work on the half-reaction for the reduction of Cr2O7 2- to Cr3+:
Cr2O7 2- (aq) → Cr3+ (aq)
We need to balance the oxygen atoms, which can be done by adding water (H2O) to the left side of the equation:
Cr2O7 2- (aq) + 7H2O (l) → Cr3+ (aq)
Now, let's balance the hydrogen atoms by adding hydrogen ions (H+) on the right side:
Cr2O7 2- (aq) + 7H2O (l) → Cr3+ (aq) + 14H+ (aq)
Finally, let's balance the charges by adding electrons (e-) on the left side:
Cr2O7 2- (aq) + 14H2O (l) + 6e- → Cr3+ (aq) + 14H+ (aq)
Multiplying the oxidation and reduction half-reactions by appropriate coefficients, we have:
6Fe2+ (aq) + 6e- → 6Fe3+ (aq)
Cr2O7 2- (aq) + 14H2O (l) + 6e- → 6Cr3+ (aq) + 14H+ (aq)
By adding these half-reactions, the electrons on both sides will cancel out:
6Fe2+ (aq) + Cr2O7 2- (aq) + 14H2O (l) → 6Fe3+ (aq) + 6Cr3+ (aq) + 14H+ (aq)
Now, looking at the balanced equation, we can see that the smallest possible integer coefficient for Cr3+ is 6. Therefore, the smallest possible integer coefficient of Cr3+ in the combined balanced equation is 6.