Sure thing!

So ur electrolytic cell uses a dc current to force a non spontaneous redox reaction.

The cathode is the sight of reduction. It’s mass will increase, as the electrons are donated to the ions, plating it in metal. Hence ‘electro-plating’

Electrons flow in the opposite direction to conventional current. They flow to the cathode. Hence the one on the right would be the cathode.

There exists a E0 for each redox reaction, which (for electrolytic cells) is the voltage needed for the reaction to occur (actually it’s anything greater than E0)

However, if the voltage gets too high, then side reactions might occur. These aren’t good (as they can make it impure/release gases and stuff)

Part D seems like a hard-to-answer question. Current is conducted through the wire I guess, and through the movement of ions in the solution? I’m not sure what they are after there…

Hope this helps!

    Thank you God

    Appreciate your help mate!

    -jinx_58

    • God likes this.

      jinx_58 for part d, maybe they're asking for the conventional current through the wire which is opposite in direction to e- flow. e-s go from anode to cathode, so you'd draw conventional current from cathode to anode since that's supposed to represent flux of positive charge. The electrolyte is part of the circuit as well, since cations are attracted to the cathode, you'd draw conventional current in the electrolyte going from anode to cathode

          8 days later

          Judging from your opinions of the chem exam this year. What would you say is the range of marks out of 120 for 47ss given rank 1 in sacs.

          2 months later

          Hi there,

          I just wanted to ask how to answer some chem questions, but one of them involves an image do I don't know how to add the link because when I don't spits out pages of code? Does anyone know how I could share the images without this issue?

            Billzene ahhh my google drive has been playing for a few days, but I'll just put up the ones that I can type:

            1. Calculate the pH of a 6.50 x 10^-3M KOH solution? (Also for this what does KOH mean??)

            2. Calculate the pH of a 0.227M solution of hydrocyanic acid (HCN). The Ka=6.2 x 10^-10 (Alsp is Ka the same as pKa.)

            3. 31.8g of Na2CO3 was dissolved to make a 500mL solution. This was diluted by pippetting 5mL of this solution into a 50mL volumetric flask and diluting to volume with deionised water. This 50mL was titrated and the results showed that the solution completely reacted with an unknown concentration of 16.8mL of H2SO4. What was the concentration of H2SO4 to 2 decimal places?

            Could anyone explain these to me in steps, thank youuu. Also I know these are all extremely simple questions, but my head has been having a hard time with readjusting to school and having anxiety (funnnn). So yeah, sorry about that everyone.

                Bibliii

                Hi 🙂

                1. KOH is potassium hydroxide, which is a base. (When in solution, it goes to OH- and K+ ions)

                So [OH-] = [KOH]

                And then [H+] = 10^-14/[OH-]

                And then ph= -log 10 ([h+]}

                (I could be slightly wrong there, I am a little rusty)

                I can’t do 17 since I did VCE (vic high school) and we didn’t have Ka. Billzene might be able to help though)

                And Q18 is giving me serious ptsd just looking at it , and i might have a crack tomorrow. 😉

                Night!
                -G

                  Bibliii
                  Hey Bibliii,

                  I think QCE does it a different way so here are the solutions to all three questions in "QCE" terms (correct me if I'm wrong):
                  16. KOH -> K+ + OH-
                  So, as @God stated, there is a 1:1 molar ratio so [OH-] = [KOH] = 6.50 x 10^-3 M
                  Since pOH = -log[OH-] = -log(6.50 x 10^-3 M) = 2.19
                  So, to find pH = 14 - pOH = 14 - 2.19 = 11.81

                  17. HCN -> H+ + CN-
                  Ka = [H+][CN-]/[HCN] = 6.2 x 10^-10 (pKa is just -log(Ka))
                  [H+][CN-]/[HCN] = (x)(x)/(0.227 - x) = 6.2 x 10^-10 and we usually assume x to be quite small so we take it as negligible (or we can calculate it with it but there's no point of wasting time for it).
                  Thus, x2/0.227 M = 6.2 x 10^-10
                  x2 = 1.407*10^-10
                  x = 1.19 x 10^-5
                  [H+] = 1.19 x 10^-5
                  pH = -log[H+] = -log(1.19 x 10^-5) = 4.93

                  18. The molar mass of Na2CO3 = 105.99 (you can calculate this through the periodic table but I searched it up from Google because I was lazy).
                  So since the number of moles (n) = mass/molar mass, the number of moles (n) of Na2CO3 = 31.8 g/105.99 g mol^-1 = 0.30 mol
                  This means concentration (c) = n/V = 0.30 mol/0.5 L = 0.60 M (concentration of Na2CO3 in 500 mL solution).
                  The dilution factor is 10 because it is from 5 mL solution into 50 mL (this can also be worked out from C1V1=C2V2) so this means the concentration of Na2CO3 in 50 mL solution = 0.06 M.
                  For the titration part, your best friend is C1V1=C2V2 so all you have to do is sub in all the values. The LHS will be the base part (Na2CO3) and the RHS is the acid part (H2SO4).
                  C1V1=C2V2
                  0.06 M x 50 mL = C2 x 16.8 mL (you should convert all mL->L like in the concentration calculation at the start but I ignored it here because it would be the same answer).
                  C2 = 0.06 x 50 / 16.8 = 0.18 M
                  Therefore, the concentration of H2SO4 is 0.18 M.

                  Make sure to always write the reaction equation at the start of the question - that's very important. This is how I was taught to calculate when I did QCE but anyone, let me know if anything's wrong because it's been a while. Hope that helps!

                  • PP

                      Bibliii KOH as God notes is the strong base potassium hydroxide, potassium is denoted by the symbol K since it’s called kalium in Latin. To approach this question, I’d take -log10( ) of the concentration of KOH they gave you to get the pOH, and then rearrange the formula pOH + pH = pKw = 14 for pH

                      For 17, hydrocyanic acid is a very weak acid judging from its low Ka magnitude so set up an ICE table as follows

                      Initial [HCN] = 0.227 M, initial [CN-] = 0 M, initial [H+] = 0 M
                      Change [HCN] = -x M, change [CN-] = +x M, change [H+] = +x M
                      Equilibrium [HCN] = 0.227 - x M, equilibrium [CN-] = x M, equilibrium [H+] = x M

                      Then equate the magnitude of Ka with your Ka expression

                      6.2 x 10^-10 = (x M)2 / (0.227 M)*

                      *In uni chem here in Victoria we’re always allowed to assume that the concentration change of an weak acid is 0, if QCE doesn’t allow it, you’ll have to use the quadratic formula.

                      Now solving for x should give you [H+], which you should then convert into pH

                      Ka isn’t the same as pKa, pKa is -log10(Ka) (whenever you see p in chem notation, it means -log10( ), such that pH = -log10(H+) and pOH = -log10(OH-))

                      18 is a titration question that involves a dilution, always start with a balanced equation and remember that your standard was diluted by a factor of 10.

                      So you know the original concentration of your standard was 31.8 g / 106 g/mol = 0.3 mol in 500 mL = 0.6 M. Diluting that by a factor 10 (use c1v1 = c2v2, where c1 = 0.6 M, v1 = 5 mL, c2 is your unknown and v2 = 50 mL). Your DILUTED concentration would be 0.06 M

                      Now you know 50 mL of a 0.06 M Na2CO3 solution reacted with 16.8 mL of H2SO4 solution. The balanced equation is pretty simple since everything is 1:1:1:1:1 (Na2CO3 + H2SO4 —> Na2SO4 + H2O + CO2). Find n(Na2CO3), which will be equivalent to n(H2SO4) according to the balanced molar ratio. Then use c = n/v to find [H2SO4], remember that v must be in L so 16.8 mL = 0.0168 L.

                        PhytoPlankton lol got beaten by you, I can confirm that you ain’t rusty and everything is correct here even though I personally would turn 50 mL into L since the concentration (0.06 mol/L) can only cancel out litres in dimensional analysis when following the c = n/v formula, and L is the most used volume unit in chem

                          Ahh thank you very much everyone for the help!!

                          6 days later

                          Hey everyone!

                          Balancing equations are kicking my ass rn and there are a few that I'm stuck on. Any help really would be greatly appreciated!

                          How would you balance this equation?
                          ___ HCI + ____ Ca(OH)2 = ______ CaCi2 + H2O

                          Thank you in advance! 🙂

                            sodacat_
                            Hey sodacat,

                            I believe you meant CaCl in the products' side of your equation so in that case, below is the balanced equation:
                            2HCI + Ca(OH)₂ = CaCl₂ + 2H₂O
                            Also, make sure you don't superscript the 2's and subscript it where you can (but that's quite insignificant 😃).

                            Hope this helps and feel free to ask more questions!

                            • PP

                                sodacat_
                                Just adding to PhytoPlankton's excellent response with an example methodology you can use when answering these types of questions.

                                What I find really helpful when balancing equations is to first balance elements other than hydrogen and oxygen (in this case calcium and chloride), then balance the hydrogens and finally balance the oxygens.

                                1. Balancing calcium and chloride
                                  We have equal numbers of calcium atoms on both sides of the equation so we don't have to do anything.
                                  There are two chloride atoms on the right hand side of the equation and one chloride atom on the left hand side. So we need to double the number of chloride atoms on the left hand side by putting a '2' in front of HCl.

                                2HCI + ____ Ca(OH)2 --> ______ CaCl2 + H2O

                                1. Balancing hydrogen
                                  We have four hydrogen atoms on the left hand side (2 from HCl and 2 from Ca(OH)2) but only two hydrogen atoms on the right hand side of the equation. Therefore we need to double the number of hydrogen atoms on the right hand side. We do this by putting a '2' in front of H2O.

                                2HCI + ____ Ca(OH)2 --> ______ CaCl2 + 2H2O

                                1. Balancing oxygen
                                  We have two oxygen atoms on the left hand side (from Ca(OH)2) and two oxygen atoms on the right hand side. Oxygen atoms are balanced so we don't need to do anything.

                                2HCI + Ca(OH)2 --> CaCl2 + 2H2O

                                Hope that helps.

                                    PhytoPlankton

                                    GreenAcorn

                                    Thank you both! Unfortunately, I did not see these responses until after I had already figured it out and got a little help from my chem teacher (flat laptops are fatal lol), but the responses were very effective in helping me to understand it all!

                                        13 days later

                                        Hiii! Data test in two days. I'm not really sure of when to use pka, pkb, kb, ka, and h30+ and oh- ions, sooo is anyone able to give a lil brief on what questions to use these in?