Unit 4 Questions MEGATHREAD!

[Vincezor]

Consider the following buffer equilibrium system:
HCN (aq) + H2O (l)  H3O+ (aq) + CN- (aq)
What is the net result of adding a small amount of HNO3 ?
A. The pH increases slightly.
B. The pH decreases slightly.
C. The [CN-] increases slightly.
D. The [HCN] decreases slightly.

Is the answer B?

Adding HNO3 increases the [H+] concentration... which pushes the equilibrium to the left? However as LCP states that it only partially opposes the change, so the H3O+ will still be higher than original, thus the pH would have decreased slightly?

Sorry... I'm so out of it. I only *really* started chem revision today haha.

[HarveyD]

i put B as well with that thought in mind
but answer says A :/

[Andiio]

i put B as well with that thought in mind
but answer says A :/

Sure the answer isn't wrong?

[HarveyD]

dont know
thats why I posted it here

[thushan]

Yup, B it is.

[HarveyD]

ah awesome
thanks for the help guys

[vea]

Something I was pondering about...
In a galvanic cell with say Ag+/Ag and Pb2+/Pb as the half cells, as the reaction occurs, would there theoretically be a temperature change in the solutions of the beaker and hence a change in pH?

[Panicmode]

in an alkaline battery.... (involving zn and mn02)

HOW THE HECK ARE YOU SUPPOSE TO DEDUCE HALF EQS OCCURING @ THE ELECTRODES?

someone please take me thru the steps

Okay, start with the easier of the two equations. We known Zn is oxidised to Zn 2+;

Zn(s) ---> Zn 2+(aq) + 2e- (Taken from your data booklet)

Since this is an alkaline cell, the Zn 2+ ions will immeaditatly react with free OH- ions in the electrolyte forming Zn(OH)2 (s) in the following equation;

Zn 2+(aq) + 2OH-(aq)  ---> Zn(OH)2 (s)


Therefore the overall equation at the anode can be summarised as;

Zn(s) + 2OH-(aq) ---> Zn(OH)2 (s) + 2e-


The second equation is more tricky to derive. You have to know that MnO2 is reduced to Mn2O3. I know this from memory (my chem teacher drilled it into us -_-) but in a reasonable question they'd tell you what species it is reduced into. Just balance the equation as if it were done under acidic conditions first


Start with just the reactant and what it is reduced to:

MnO2(s) ---> Mn2O3(s)

Start by balancing non-oxygen/hydrogen atoms.

2MnO2(s) ----> Mn2O3(s)

Balance out oxygens with addition of water:

2MnO2(s) ----> Mn2O3(s) + H2O(l)

Balance out Hydrogens with addition of H+:

2MnO2(s) + 2H+(aq) ----> Mn2O3(s) + H2O(l)

Now balance out the charges on both sides through adding e- to the more positive side:

2MnO2(s) + 2H+(aq) 2e- ----> Mn2O3(s) + H2O(aq)


This is a nice balanced equation, but it's under acidic conditions. We need to convert it to alkaline conditions. So take the following steps:

Add as many OH- ions to both sides as there are H+ ions.

2MnO2(s) + 2OH-(aq)  + 2H+(aq) + 2e- ----> Mn2O3(s) + H2O(aq) + 2OH-(aq)

Remember that OH- ions and H+ ions will react to form water. Therefore:

2MnO2(s) + 2H2O(l) + 2e- ----> Mn2O3(s) + H2O(l) + 2OH-(aq)

There are waters on both sides of the equation, cancel these out.

2MnO2(s) + H2O(l) + 2e- ----> Mn2O3(s) + 2OH-(aq)

Your final equation for the reaction at the cathode is now complete:

2MnO2(s) + H2O(l) + 2e- ----> Mn2O3(s) + 2OH-(aq)

[HenryP]

Something I was pondering about...
In a galvanic cell with say Ag+/Ag and Pb2+/Pb as the half cells, as the reaction occurs, would there theoretically be a temperature change in the solutions of the beaker and hence a change in pH?

Yea you're right but if we are to assume that the reaction is at standard conditions, we assume that the temperature is always at 298K. Its the same issue when dealing with equilibrium questions. Say we decrease the pressure of a vessel and that pushes the exothermic equation forward, there would be a temperature increase which would push the reaction back. We just simplify all these situations by assuming temperature is kept constant unless told otherwise.

[vea]

Something I was pondering about...
In a galvanic cell with say Ag+/Ag and Pb2+/Pb as the half cells, as the reaction occurs, would there theoretically be a temperature change in the solutions of the beaker and hence a change in pH?

Yea you're right but if we are to assume that the reaction is at standard conditions, we assume that the temperature is always at 298K. Its the same issue when dealing with equilibrium questions. Say we decrease the pressure of a vessel and that pushes the exothermic equation forward, there would be a temperature increase which would push the reaction back. We just simplify all these situations by assuming temperature is kept constant unless told otherwise.

ty henry
hope you get your 30 in chem

[HenryP]

Thanks man, I've been working hard all year to get it .

[Zebra]

converting an acidic half equation in a acidic fuel cell by simply adding oh- will work always?

[Panicmode]

converting an acidic half equation in a acidic fuel cell by simply adding oh- will work always?

Yup. =D It's how we were taught to balance equations in alkaline conditions.

[david10d]

i don't really understand the solutions for this question:
Half equations for the cells:

CH4 + 8OH- -> CO2+ 6H2O + 8e-

2O2 + 4H2O + 8e- -> 8OH-

When set up, the electrolyte contained 0.5mol of dissolved OH-. The cell operates for a period of time during which 0.1mol of CH4 gas is consumed. What amount of OH-, in mol, is now dissolved in the electrolyte? Explain how you arrived at your answer.

[Asx4Life]

in an alkaline battery.... (involving zn and mn02)

HOW THE HECK ARE YOU SUPPOSE TO DEDUCE HALF EQS OCCURING @ THE ELECTRODES?

someone please take me thru the steps

Okay, start with the easier of the two equations. We known Zn is oxidised to Zn 2+;

Zn(s) ---> Zn 2+(aq) + 2e- (Taken from your data booklet)

Since this is an alkaline cell, the Zn 2+ ions will immeaditatly react with free OH- ions in the electrolyte forming Zn(OH)2 (s) in the following equation;

Zn 2+(aq) + 2OH-(aq)  ---> Zn(OH)2 (s)


Therefore the overall equation at the anode can be summarised as;

Zn(s) + 2OH-(aq) ---> Zn(OH)2 (s) + 2e-


The second equation is more tricky to derive. You have to know that MnO2 is reduced to Mn2O3. I know this from memory (my chem teacher drilled it into us -_-) but in a reasonable question they'd tell you what species it is reduced into. Just balance the equation as if it were done under acidic conditions first


Start with just the reactant and what it is reduced to:

MnO2(s) ---> Mn2O3(s)

Start by balancing non-oxygen/hydrogen atoms.

2MnO2(s) ----> Mn2O3(s)

Balance out oxygens with addition of water:

2MnO2(s) ----> Mn2O3(s) + H2O(l)

Balance out Hydrogens with addition of H+:

2MnO2(s) + 2H+(aq) ----> Mn2O3(s) + H2O(l)

Now balance out the charges on both sides through adding e- to the more positive side:

2MnO2(s) + 2H+(aq) 2e- ----> Mn2O3(s) + H2O(aq)


This is a nice balanced equation, but it's under acidic conditions. We need to convert it to alkaline conditions. So take the following steps:

Add as many OH- ions to both sides as there are H+ ions.

2MnO2(s) + 2OH-(aq)  + 2H+(aq) + 2e- ----> Mn2O3(s) + H2O(aq) + 2OH-(aq)

Remember that OH- ions and H+ ions will react to form water. Therefore:

2MnO2(s) + 2H2O(l) + 2e- ----> Mn2O3(s) + H2O(l) + 2OH-(aq)

There are waters on both sides of the equation, cancel these out.

2MnO2(s) + H2O(l) + 2e- ----> Mn2O3(s) + 2OH-(aq)

Your final equation for the reaction at the cathode is now complete:

2MnO2(s) + H2O(l) + 2e- ----> Mn2O3(s) + 2OH-(aq)

WOAH, My school never taught this, what hacks. -___-