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Chemistry

I think a maybe

Answer is D. Only equation where the total charges on both sides of the equation are equal. Or otherwise you should know the reduction of dichromate has six electrons, so you’ll need 6 iron species to balance this.

(edited 4 years ago)

Reply 3

A-Conquer

6e-+ Cr2O72- + 14H+ -> 2Cr3+ +7H20
Fe2+ +e- -> Fe3+

so D

Reply 4

_jess_here__

Original post by A-Conquer

6e-+ Cr2O72- + 14H+ -> 2Cr3+ +7H20
Fe2+ +e- -> Fe3+

so D

can u explain what happnd here?

Reply 5

_jess_here__

Original post by Deggs_14

explain?

(edited 4 years ago)

Reply 6

Ðeggs

Original post by _jess_here__

can u explain what happnd here?

The oxidation of iron (II) to iron (III) has one electron. So this equation is multiplied by 6 in order to have the number of electrons equal to the reduction of dichromate half equation. Then add the equations together and you have the full half equation as in answer D.

Otherwise, add up the total ionic charges on each side of the equation, and they only are equal in D.

(edited 4 years ago)

Reply 7

A-Conquer

Original post by _jess_here__

can u explain what happnd here?

Pretty much, you start with these two half equations
Cr2O72- > Cr3+
and
Fe2+ > Fe3+
Balance them both into what I had above and then use multiples

So the Cr2o72- equation had 6e- in it, the Iron had 1e- so we had to times the Iron equation by 6 to get the answer of D when we combine the two fully balanced half equations