CHEMISTRY COMMUNITY

Gwen Peng 1L

To solve you need the 2 half reactions: O3 + H2O + 2e- --> O2 + 2OH- E standard= 1.24 V O3 + 2H+ + 2e- --> O2 + H2O E standard= 2.07 V Your cathode is always the oxidizing agent since it undergoes reduction and anode is always the reducing agent since it undergoes oxidation. Your anode is always the...

Gwen Peng 1L

n refers to the number of electrons moved, or electrons on the product side, therefore since 4 electrons were transferred n=4

Gwen Peng 1L

The final most likely had a sheet with all of the standard cell potentials which then allows you to determine your anode and cathode. The anode is always the half reaction that has a more positive standard cell potential (higher voltage) because it will be flipped and the sign will become negative, ...

Gwen Peng 1L

You flip the anode (which should have the higher voltage) because then when you plug into the equation E standard (cell) = E standard (cathode) - E standard (anode) it will be a double negative which then becomes a positive and results in a positive overall standard cell potential. Hope this helps, ...

Gwen Peng 1L

For the reaction: Mn^2+ + Br2 --> MnO4^- + Br^- In order to balance in acidic conditions you would need to add H2O to the reactants side and H+ to the product side to get: 4H2O + Mn^2+ + 5Br2 --> MnO4^- + 5Br^- + 8H+ And then the equation is balanced, but the H+ is on the products side rather than t...

Gwen Peng 1L

For this problem it says the half-life of A is 50.5 s when [A]0= 0.84 mol/L. It asks to calculate the time needed for the concentration of A to decrease to a) 1/16; b) 1/4; c) 1/5. I understand you would use t=\frac{\frac{1}{[A]}-\frac{1}{[A]_{0}}}{k} however in each part it solves like this: a) t=\...

Gwen Peng 1L

If you're given the 2 half reactions and their standard cell potentials for a couple do you always make the reaction with the more positive standard cell potential the anode because when you flip it it will become negative and when plugged into the equation it still makes the overall standard cell p...

Gwen Peng 1L

RenuChepuru1L wrote:are you sure this was test one?? we hadn't gotten to reduction in test one yet, it was only chapter 8 and reduction stuff is in chapter 14

My fault it was a typo, but yes the same question only for test 2 not test 1

Gwen Peng 1L

RenuChepuru1L wrote:I think where you're getting confused is E standard= Ecathode-Eanode so if you "flip" the sign you wouldn't subtract again, you'd add

So since there is a double negative it would be E standard= 1.69 V + 0.34 V, correct?

Gwen Peng 1L

904940852 wrote:For this one Q= [H+]^s[HCl]^2/[Pressure of H2]

Can you please explain why you would use this?

Gwen Peng 1L

I don't believe so because E= E standard - (0.05916/n)*log(K) and if E= 0 then you would move E standard to the other side to have -E standard= -(0.05916/n)*log(K) and the negatives would cancel.

Gwen Peng 1L

This question asks to find the standard cell potential of the following: Cu|Cu2+||Au+|Au It gives the half reactions: Cu2+ + 2e- --> Cu E standard= 0.34V Au+ + 1e- --> Au E standard= 1.69 V As shown in the cell diagram the Au is the cathode and the Cu is the anode, but you would have to flip the hal...

Gwen Peng 1L

For this question it gave the reaction:
2Ag + H2S --> Ag2S + H2+
Then it asked which species is reduced.
The answer is H2, but I was wondering why it is just H2 and not H2S?

Gwen Peng 1L

It says that I- is the anode and Cl- is the cathode and gives the following:
[I-] = 0.025 M
[Cl-] = 0.67 M
Then it asks to calculate Q. Wouldn't you just use Q= [anode]/[cathode]= [0.025]/[0.67] to calculate?
I am asking because this is what I did, but my answer was marked wrong.

Gwen Peng 1L

For this question the galvanic cell is formed with 03/02, OH- and )3, H+/O2 and it gave the following half reactions:
O3 + H2O + 2e- --> O2 + 2OH-
O3 + 2H+ + 2e- --> O2 + H2O
If the oxidizing agent (cathode) is O3 then what is the reducing agent (anode)?

Gwen Peng 1L

Yes it was assumed it was in equilibrium so you could've solved using the equation E standard=(0.05916/n)*log(K) or E standard=(RT/nF)*ln(K)
Hope this helps :)

Gwen Peng 1L

If you are given 2 half reactions and their standard cell potentials then you are asked to find the standard cell potential of the combined reaction, why do you have to find the Gibbs Free Energy of each half reaction and combine them to find the standard cell potential of the reaction rather than j...

Gwen Peng 1L

Is balancing a reaction under basic conditions the same process as solving for acidic only you take it one step further by adding OH-? So it would be the following: Step 1: Identify half reactions (which is undergoing oxidation/reduction) Step 2: Balance the atoms and charges on each half reaction S...

Gwen Peng 1L

Wenjie Dong 2E wrote:E_cell=-0.44 V-(-0.74 V)=0.30V, and then use the equation mentioned above to calculate K.

Could you explain why you used -0.44 V as the cathode value if Fe is being oxidized since it gains electrons? Shouldn't -0.74 V be the cathode value since Cr is being reduced since it lost electrons?

Gwen Peng 1L

For this question wouldn't you just use the formula: E(cell)=E(cathode)-E(anode), which means it would be +2.07 V- (-1.24 V)= +3.31 V?

Gwen Peng 1L

Can someone please explain this problem? It asks to find K of Cr^3+(aq)+Fe(s) -> Fe^2+(aq)+Cr(s) and gives the following:
Fe^2+(aq)+2e- -> Fe(aq) E=-0.44 V
Cr^3+(aq)+3e- -> Cr(s) E=-0.74 V

Gwen Peng 1L

How do you determine which elements have a higher reducing power?

Gwen Peng 1L

Can someone please explain this problem? It gives the standard cell potential for Cr2O7^2-+14H^+ + 6e- -> 2Cr^3++7H2O is 1.33 V, then asks to find the standard cell potential of the same reaction but with 12e- (double the electrons).

Gwen Peng 1L

On cell diagrams do we always add Pt when there is a group of elements separated by a comma? And would we add Pt when elements are drawn with |, which represents contact, or only when there is a comma?
Also, how do we know when to use a comma and when to use a |?

Gwen Peng 1L

On Test 2 Question 6 Part C, we drew the cell diagram with the O3 and O2 together but separated by a comma, did we group them together since they are the same element in one reaction even though one is a reactant and one is a product? If so, should we always group elements together in cell diagrams?

Gwen Peng 1L

When drawing cell diagrams, do you always put it in the following order?
(Anode reactant)|(Anode product)||(Cathode reactant)|(Cathode product)

Gwen Peng 1L

First, I found the two half reactions on the list provided at the back of the test. I then flipped the reaction with the lower voltage (making it an oxidation half reaction because there must be one oxidation and one reduction) and that told me that the O3/O2, OH- is the oxidation half reaction. An...

Gwen Peng 1L

The "en-" in endothermic sounds like enter, as in heat is entering, and the "ex-" in exothermic sounds like exit, as in heat is exiting.

Gwen Peng 1L

The extensive properties scale directly with size, i.e. if the size of a system doubles, the value of an extensive property simply doubles as well. Intensive properties, on the other hand, would simply remain constant, whether the system size is doubled, tripled, or changed in any way. This distinct...

Gwen Peng 1L

Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes.

Gwen Peng 1L

Yes, in a combustion equation the reactants will always be whatever compound is burning and oxygen (O2) and the products will always be carbon dioxide (CO2) and water (H2O).

Gwen Peng 1L

A limiting reactant determines the amount of product created from a reactant. This means that a reaction will stop once the limiting reactant in the reaction is completely used up. Whereas, the theoretical yield is an estimation of the amount of product which is predicted using a stoichiometric calc...

Gwen Peng 1L

So yes the mass of a neutron is always constant. The mass of a neutron is 1 amu. Therefore you convert this to kilograms and get:
1 amu = 1.67493x10^-27 kg
Hope this was helpful :)

Gwen Peng 1L

Yes in order to solve this problem you must convert to joules, so it should be: E= (140.511x10^3 eV)(1.0622x10^-19 J*eV^-1) = 2.2513x10^-14 J Then you can plug this value into the following equation: wavelength = (hc)/E = ((6.626x10^-34 J*s)(3.00x10^8 m*s^-1))/2.2513x10^-14 J = 8.8237x10^-12 m or 8....

Gwen Peng 1L

You are correct and CaCO3 is the limiting reactant. To get the theoretical yield it would be: (42.73 g CaCO3) x ((1mol CaCO3)/100.09 g CaCO3) x ((1 mol CO2)/1 mol CaCO3) x ((44.01 g CO2)/ 1 mol C02) = 18.79 g CO2 To get percent yield just divide the actual yield by the theoretical yield and multiply...

Gwen Peng 1L

To begin you would write the two separate chemical equations, like this: 4.84 g MgO + CaO and 9.66 g MgCO3 + CaCO3 Next you just sub in x and y for the moles of each: x mol MgCO3 and y mol CaCO3 Then plug these into two separate equations using the molar masses of each compound: x(M(MgCO3)) + y(M(Ca...

CHEMISTRY COMMUNITY

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