Which element is reduced in the reaction below? Fe(CO)5 (I) + 2HI (g) rightarrow

Question

Which element is reduced in the reaction below? Fe(CO)5 (I) + 2HI (g) rightarrow Fe(CO)412(s) + CO(g) + H2(g) A) Fe B) C C) O D) H E) I 2) What is the coefficient of the dichromate ion when the following equation is balanced? Fe2+ + Cr2O72- rightarrow Fe3+ + Cr3+ (acidic solution) A) 1 B) 2 C) 3 D) 5 E) 6 3) Which transformation could take place at the anode of an electrochemical cell? A) Cr2O72- rightarrow Cr2+ B) F2 to F- C) O2 to H2O D) HAsO2 to As E) None of the above could take place at the anode. 4) The purpose of the salt bridge in an electrochemical cell is to. A) maintain electrical neutrality in the half-cells via migration of ions B) provide a source of ions to react at the anode and cathode C) provide oxygen to facilitate oxidation at the anode D) provide a means for electrons to travel from the anode to the cathode E) provide a means for electrons to travel from the cathode to the anode.

Explanation / Answer

I will answer 1 and 2. Try to answer question 3 yourself with the explanation from 1 and 2. It's not that hard.

Question 1. to know the element that is reduced, you need to calculate the oxidation state of all elements in the reaction and compare the changes. If you go from a high number to a lower number (from 2+ to 0, or -1 to -3), it's a reduction. It's called reduction because the element is gaining electrons (when an element gets positive means that it lacks of electrons).

Now in this reaction:
Fe(CO)5 + 2HI -----> Fe(CO)4I2 + CO + H2

We can see that Carbon is in the CO, and it's remain the same in the products, so it oxidation state should be the same. in the molecule of HI, hydrogen usually works with an oxidation state of 1+, which means that Iodine should be working with 1-. In the products, we see that Hydrogen changes to H2, and when we have diatomic compounds, or solid atoms, the oxidation state is always zero (0). so in this case H2 is 0, and change from 1+ to 0, which means that it gained one electron, and this element was reduced.

2. To know this, balance the half reactions:
(Fe2+ -------> Fe3+ + 1e-)*6
14H+ + Cr2O72- + 6e- ------> 2Cr3+ + 7H2O

6Fe2+ -------> 6Fe3+ + 6e-
14H+ + Cr2O72- + 6e- ------> 2Cr3+ + 7H2O

6Fe2+ + Cr2O72- + 14H+ --------> 6Fe3+ + 2Cr3+ + 7H2O

The coefficient is 1.

For question 4, usually a salt bridge, is like that a bridge that connects both half cells. Salt bridge allows electrons to be transfered, and the element that is reduced, is the element that is gaining electrons, so the salt bridge allows electrons to travel from anode (oxidation) to cathode (reduction).

Now try to answer question 3. Remember that anode is oxidation (lose electrons) and cathode is reduction (gains electrons).

Hope this helps

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