Balance each of the following skeletal equations by using oxidation and reduction half reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
(a) Conversion of iron(II) into iron(III) by dichromate ion
Fe2+(aq) + Cr2O72-(aq) → Fe3+(aq) + Cr3+(aq)
For the equation, I put this and it returned:
"Your answer appears to be missing one or more states-of-matter."
6Fe^2+ + Cr_2O_7^2- + 14H^+ --> 6Fe^3+ + 2Cr^3+ + 7H_20
For the reducing reagent I put this and it returned: "Your answer appears to substitute one element for another."
Cr^6+
For the oxidizing reagent I put this and it returned: "Your answer differs too much from the expected answer to provide useful feedback."
Can someone help me correct what I did wrong?
Sorry, for the oxidizing, I put this: Fe^2+
For the equation you didn't add any of the states; i.e., 6Fe^2+(aq) + Cr2O7^2-(aq) etc
Cr2O7^2-(aq) is the oxidizing agent; Fe^2+(aq) is the reducing agent.
To balance the equation, you have attempted to balance the atoms and charges correctly, but you forgot to include the states-of-matter under SATP (Standard Ambient Temperature and Pressure) conditions.
The corrected balanced equation, including states-of-matter under SATP conditions, would be:
6Fe2+(aq) + 2Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
Now, let's identify the oxidizing and reducing agents in this reaction.
The reducing agent is the species that is oxidized, meaning it loses electrons. In this case, Fe2+ is oxidized to Fe3+, so Fe2+ is the reducing agent.
The oxidizing agent is the species that is reduced, meaning it gains electrons. In this case, Cr2O72- is reduced to Cr3+, so Cr2O72- is the oxidizing agent.
Therefore, the reducing agent is Fe2+ and the oxidizing agent is Cr2O72-.