2012 HSC chemistry Multiple Choice (1 Viewer)

[DaLaBa]

Hey does anyone know how to do this?

This equation represents a common redox reaction.
Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
14 What is the value of E−cell ° for the reaction?
(A) 0.59 V
(B) 0.92 V
(C) 1.90 V
(D) 2.13 V

The answer is A
Thanks

 

[Sp3ctre]

My chemistry knowledge is a bit rusty, but hopefully this makes sense.

By calculating the oxidation numbers of Cr and Fe in the equation, you should see that Cr goes from a charge of 6+ to 3+ while Fe goes from charge 2+ to 3+, meaning that Cr is gaining electrons and therefore undergoing reduction, whereas Fe is losing electrons and undergoing oxidation. Look at your standard reduction table on the data sheet and find the half equations for Cr and Fe and find their reduction potentials. Fe is 0.77 and Cr is 1.36. Since Fe is oxidising, just change the sign to get its oxidation potential, which would be -0.77. Adding these together should give you the overall standard cell potential.

 

[DaLaBa]

My chemistry knowledge is a bit rusty, but hopefully this makes sense.

By calculating the oxidation numbers of Cr and Fe in the equation, you should see that Cr goes from a charge of 6+ to 3+ while Fe goes from charge 2+ to 3+, meaning that Cr is gaining electrons and therefore undergoing reduction, whereas Fe is losing electrons and undergoing oxidation. Look at your standard reduction table on the data sheet and find the half equations for Cr and Fe and find their reduction potentials. Fe is 0.77 and Cr is 1.36. Since Fe is oxidising, just change the sign to get its oxidation potential, which would be -0.77. Adding these together should give you the overall standard cell potential.

Im still a little confused because the Cr standard potential has +3e- while the Fe has +e-. Dont you have to times the Fe by three so it is balanced out? Thanks.

 

[Sp3ctre]

Im still a little confused because the Cr standard potential has +3e- while the Fe has +e-. Dont you have to times the Fe by three so it is balanced out? Thanks.

You never multiply the cell potential, only multiply whatever is in the half equation when your trying to make a full equation from the two half equations, which you don’t need to in this q anyway

 

[DaLaBa]

You never multiply the cell potential, only multiply whatever is in the half equation when your trying to make a full equation from the two half equations, which you don’t need to in this q anyway

Oh ok. Thanks for your help

 

[jazz519]

Oh ok. Thanks for your help

We don't need to multiply the voltage value by 3, because voltage refers to a potential difference between the two cells which just relates to how reactive they are. If it was talking about current formed then in that case we would multiply by 3 since current relates to the amount of electrons, but note that isnt covered in the HSC so its fine to just remember molar ratios have no effect on voltage.