F325 -balancing

hi can someone please show me how to balance this

Zn gives Zn(2+) + e-
Cr2o7(2-) + H+ + e- gives Cr(2+) + H20
the brackets are the charges

ok i know how to balance the first one
you do
Zn gives Zn(+2) + 2e-

then second part i balanced everything except the electron how do u balance the electron they got 8 ? how?

Reply 1

otrivine

and should there have to be a 2 in front of cr2o7(2-)?

Reply 2

Dreamweaver

Ok, this is how you form and balance half equations:

First off, learn the basic first step. Dichromate ions are reduced to Chromium ions:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- \to \ 2(Cr)^3^+

Oxygen always comes out as water:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- \to \ 2(Cr)^3^+ +7H_2O

We need more hydrogens on the LHS to balance the equation so add H+:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- +14H^+ \to \ 2(Cr)^3^+ +7H_2O

Now balance the electrons:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- +14H^++6e^- \to \ 2(Cr)^3^+ +7H_2O

(edited 11 years ago)

Reply 3

otrivine

Original post by Dreamweaver

Ok, this is how you form and balance half equations:

First off, learn the basic first step. Dichromate ions are reduced to Chromium ions:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- \to \ 2(Cr)^3^+

Oxygen always comes out as water:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- \to \ 2(Cr)^3^+ +7H_2O

We need more hydrogens on the LHS to balance the equation so add H+:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- +14H^+ \to \ 2(Cr)^3^+ +7H_2O

Now balance the electrons:

Unparseable latex formula:

\displaystyle (Cr_2O_7)^2^- +14H^++6e^- \to \ 2(Cr)^3^+ +7H_2O

its wrong the electrons has to be 8?

Reply 4

Dreamweaver

Original post by otrivine

its wrong the electrons has to be 8?

No it's not wrong. If you add up the charges you will find it is:

+6 -> +6

Therefore the charges are balanced.

Reply 5

otrivine

Original post by Dreamweaver

No it's not wrong. If you add up the charges you will find it is:

+6 -> +6

Therefore the charges are balanced.

http://pdf.ocr.org.uk/download/ms_10/ocr_56094_ms_10_gce_f325.pdf?

mark scheme question 6)b) check

Reply 6

kurdishboy94

Original post by otrivine

http://pdf.ocr.org.uk/download/ms_10/ocr_56094_ms_10_gce_f325.pdf?

mark scheme question 6)b) check

na what he is saying is right you have 8 electrons for the second equation then u multiply the zinc equation by 4 hence u get 8 electrons i dunno where u got 6, i understand the over equation needs 6 electrons plus 2- from the cr2072- you get 4+ on both sides hence an equal and balanced equation.

check it if u dont believe me.

Reply 7

Dreamweaver

Original post by otrivine

http://pdf.ocr.org.uk/download/ms_10/ocr_56094_ms_10_gce_f325.pdf?

mark scheme question 6)b) check

Original post by kurdishboy94

Both equations are correct. On the MS, the Cr ion has a 2+ charge but in the book you will find examples where Cr is 3+. This is because Cr can have different oxidation states. You need to be careful and always look at the question.