how many electrons are transferred in the balanced reduction half-reaction?

Question

Cr₂O₇^2–(aq) + NO₂^–(aq) → Cr³⁺(aq) + NO₃^–(aq)

If the reaction occurs in an acidic solution where water and H⁺ ions are added to the half-reactions to balance the overall reaction, how many electrons are transferred in the balanced reduction half-reaction?

Answer 1

Barry M has given a cogent and correct answer, but I thought I'd present method 2 in a stepwise, outline approach to hopefully make it easier to see what is being done/accomplished. We will deal with the 2 half reactions separately.

Cr₂O₇^2- ==> Cr³⁺ ... reduction half reaction because Cr goes from 6+ to 3+, gaining electrons

Cr₂O₇^2- ==> 2Cr³⁺ ... to balance the Cr atoms

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O ... to balance the O atoms

Cr₂O₇^2- + 14H⁺ ==> 2Cr³⁺ + 7H₂O ... to balance the H atoms using acid

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O ... to balance the charge

NO₂^- ===> NO₃^- ... oxidation half reaction because N goes from 3+ to 5+, losing an electron

NO₂^- + H₂O ===> NO₃^- ... to balance the O atoms

NO₂^- + H₂O ===> NO₃^- + 2H⁺ ... to balance the H atoms using acid

NO₂^- + H₂O ===> NO₃^- + 2H⁺ + 2e- ... to balance the charge

You would then multiply this final oxidation half reaction by 3 (x3) to equalize the electrons and SIX electrons would be transferred.

As stated, the original answer is fine, and correct. Just wanted to offer an additional setup.

Answer 2

Let's look at 2 approaches to this question.

1. In the Cr₂O₇^2- ion, the Cr atoms each have a charge of 6+. That is calculated by the standard assignment of 2- to each O atom, and the need for all the charges to add up to 2-. Clearly, 7 O atoms at 2- each makes 14-, so 2 Cr atoms must add up to 12, in order to give the total of 2- for the dichromate ion as given. The product Cr is given as 3+. So, from the reactant Cr having 6+ to the product Cr having 3+, there is a reduction of 3 for each Cr.

The question requires the overall reaction being balanced--all redox reactions have oxidation and reduction occurring simultaneously.

In order for the overall reaction to be balanced, it will be necessary for the oxidation to gain the same charge as is lost by reduction. Looking at the oxidation, N in the nitrite ion reactant has a charge of 3+ (as above, O is assigned 2-, so to achieve the given 1-, N must be 3+, as the 2 O atoms will have 4-), and N in the nitrate ion product has 5+ (since the 3 O atoms have 6- and nitrate is 1-). Therefore, for each N in the oxidation, the charge is increasing by 2.

In the balanced equation, the charge increase by oxidation must equal the decrease by reduction. This is achieved by multiplying each reaction by the lowest appropriate integer that will achieve that goal. In this case, The Cr reaction would be multiplied by 2, and the N by 3, so the transfer of charge would be 6 in each half reaction. Note that electrons have a charge of 1- each, so 6 electrons are gained in reduction, and 6 electrons are lost in oxidation.

2. In this method, balance all the atoms in each half reaction, using H₂O to balance the O atoms as needed, and using H+ to balance H as needed. Also, balance the total charge on each side of the half reaction by adding an appropriate number of e^-, each of which has a 1- charge. In this problem, there will be a half reaction for Cr and one for N.

Balance the Cr by giving a coefficient of 2 to Cr in the products, matching the 2 Cr atoms in each dichromate ion. You will need 7H₂O in he products to match the 7 O atoms in each dichromate. This will then require 14H⁺ in the reactants to balance the H atoms in 7H₂O. Finally, add 6e^- to the reactants in order to balance the total charge of 6 on both sides of the reaction.

Next, balance N by adding one H₂O, to the reactants, requiring 2H⁺ and 2e^- to the products. Since the e^- transfer in oxidation must be balanced by that of reduction, all the coefficients (of course, when no coefficient is explicitly written, it is assumed to be 1, by convention) in this reaction will need to be multiplied by 3, giving a total of 6e^- in the products.

So it would appear that by using either method 1 or method 2, there will be a transfer of 6 electrons in each balanced half reaction. By the way, it is common for the question to require balancing of the complete reaction, so that would be obtained by adding the half reactions together (directly from method 2, or with some additional work following method 1). In the process, remember to cancel out excess H⁺ or H₂O from appropriate sides of the equation. If the work was done properly, the e^- on each side will automatically be equal and completely cancel.