Question

Select the correct balanced oxidation-reduction reaction. Group of answer choices 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr +(aq) + 7H2O(l) 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 2e- 7H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Answer 1

Answer:

The correct balanced oxidation- reduction reaction is:

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Explanation:

In this reaction, iron (Fe2+) is the reducing agent while Cr2O7^2- is the oxidizing agent.

The ion transfer is represented as shown below:

6 Fe 2+  - 6e- -----------> 6 Fe 3+          (oxidation)

2 Cr^6  + 6e^-  ----------> 2 Cr^3           (reduction)

From the unbalanced reaction

Fe2+ + Cr2O72- + H+ ---------->  Fe3+ + Cr3+ + H2O we will follow these steps to balance the reaction.

Step 1: break the equation into two half reactions stating which is oxidized and reduced.

Step 2: Balance the atoms on each sides; the hydrogen, oxygen

Step 3: Balance the gain also

Step 4: Give the electron gained on one side to be equal to the electron lost on the other side.

Step 5: Add the two half reactions and simplify the equation.

Doing this, we obtain

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)