Chem 08 sample paper Q

[samiira]

In the last bit of the answer it says 'twice as much nickel will be produced.' so shouldnt the answer be d..??

[jasoN-]

I think they meant 'twice as much silver will be produced'.
As we know Ag+ + e -> Ag and Ni2+ + 2e -> Ni, Nickel requires twice as much electrons.
As it's given a constant current throughout the system, more Ag will be produced in a 2:1 ratio (ie. 1 mol of electrons produces 1 mol of Ag and 0.5 mol of Ni)

[samiira]

kool..thanks

[samiira]

Q7 ci and ii from the same paper.. how do u find the equation for the cathode and overall eq for this q.

http://www.vcaa.vic.edu.au/vce/studies/chemistry/2008chem2-samp.pdf

[scocliffe09]

Q7 ci and ii from the same paper.. how do u find the equation for the cathode and overall eq for this q.

http://www.vcaa.vic.edu.au/vce/studies/chemistry/2008chem2-samp.pdf

Question says "hydrogen gas and oxygen gas as the reactants" and "water is the only product" and you are given the anode half equation.

So overall is 2H2+O2-->2H2O
to get cathode half cell, just "subtract" the anode equation from the full equation...
i.e. O2+2e- --> 2O2-

[samiira]

Question says "hydrogen gas and oxygen gas as the reactants" and "water is the only product" and you are given the anode half equation.

So overall is 2H2+O2-->2H2O
to get cathode half cell, just "subtract" the anode equation from the full equation...
i.e. O2+2e- --> 2O2-

thanks alot..but i got confused with ur last part.. isnt it meant to be 4e- instead of 2e-..??

[scocliffe09]

yes sorry 4e !

[matt123]

Q7 ci and ii from the same paper.. how do u find the equation for the cathode and overall eq for this q.

http://www.vcaa.vic.edu.au/vce/studies/chemistry/2008chem2-samp.pdf

How did you work out the overall equation?
Question says "hydrogen gas and oxygen gas as the reactants" and "water is the only product" and you are given the anode half equation.

So overall is 2H2+O2-->2H2O
to get cathode half cell, just "subtract" the anode equation from the full equation...
i.e. O2+2e- --> 2O2-

[jasoN-]

An Internet site reporting the latest developments in fuel cell technology describes a cell that uses a solid ceramic
material as the electrolyte and hydrogen gas and oxygen gas as the reactants.
Key features of this cell are
• water is the only product from the cell reaction
• the ceramic material allows the movement of oxide ions (O2–)
• the reaction at the anode is H2(g) + O2–(in ceramic) → H2O(l) + 2e–
• operation at very high temperatures of over 1000°C means that precious metal catalysts are not required.
A representation of the cell providing electricity for an appliance is shown in the diagram below.

indicative of H2 and O2 as the reactants, H2O the product;
therefore; 2H2 + O2 -> H2O

With the anode equation given, simply derive the cathode by subtracting reactants

[matt123]

An Internet site reporting the latest developments in fuel cell technology describes a cell that uses a solid ceramic
material as the electrolyte and hydrogen gas and oxygen gas as the reactants.
Key features of this cell are
• water is the only product from the cell reaction
• the ceramic material allows the movement of oxide ions (O2–)
• the reaction at the anode is H2(g) + O2–(in ceramic) → H2O(l) + 2e–
• operation at very high temperatures of over 1000°C means that precious metal catalysts are not required.
A representation of the cell providing electricity for an appliance is shown in the diagram below.

indicative of H2 and O2 as the reactants, H2O the product;
therefore; 2H2 + O2 -> H2O

With the anode equation given, simply derive the cathode by subtracting reactants

Yeah
I dont get this 02- Thing though.
The cathode equation also shows 2O2- ? wat the hell is that lol.
Im still having trouble getting the FULL equation..

grrr i hate this stuff

[matt123]

I JUST REALISED H + 02 ONLY REACTANTS

lol dw i think i get it now
thanks.