Redox Calculations (1 Viewer)

[Dreamerish*~]

A beaker initially contained 250.0 mL of 0.050 mol L^-1 copper sulfate solution. After several hours, the dark blue colour of the solution had become lighter and a red-brown deposit had formed on the piece of zinc metal. The red-brown deposit was removed from the piece of zinc metal and dried. It was found to weigh 0.325 g. Calculate the concentration of copper sulfate solution remaining in the beaker.

Please do it and tell me if you get 0.030 mol L^-1 (3 d.p.)

 

[Haku]

A beaker initially contained 250.0 mL of 0.050 mol L^-1 copper sulfate solution. After several hours, the dark blue colour of the solution had become lighter and a red-brown deposit had formed on the piece of zinc metal. The red-brown deposit was removed from the piece of zinc metal and dried. It was found to weigh 0.325 g. Calculate the concentration of copper sulfate solution remaining in the beaker.

Please do it and tell me if you get 0.030 mol L^-1 (3 d.p.)

moles of Cu in solution is (0.05/1000)*250. = 0.0125
mass of cu is 0.0125*63.55 = 0.794375
mass left is 0.794375 - 0.325 = 0.469375

this much in 250ml is 1.8775g in 1 litre
which is 0.02954....moles per litre.

guess thats right.

oh and just asking what year hsc paper is that question from?

 

[Dreamerish*~]

moles of Cu in solution is (0.05/1000)*250. = 0.0125
mass of cu is 0.0125*63.55 = 0.794375
mass left is 0.794375 - 0.325 = 0.469375

this much in 250ml is 1.8775g in 1 litre
which is 0.02954....moles per litre.

guess thats right.

oh and just asking what year hsc paper is that question from?

Oh good!

The "notes" only come with stupid useless remarks and the bleeding obvious, but no answers.

It's from 2004.

 

[Haku]

haha, i am kinda stuk for part a) of that question

for the equation what do i write?

 

[Dreamerish*~]

haha, i am kinda stuk for part a) of that question

for the equation what do i write?

For redox reactions, writing half equations is very very important. It is equally important to shove your states in.

The oxidation half-equation is: Zn_(s) → Zn²⁺_(aq) + 2e^-

The reduction half-equation is: Cu²⁺_(aq) + 2e^- → Cu_(s)

The overall reaction is: CuSO_4(aq) + Zn_(s) → ZnSO_4(aq) + Cu_(s)

Take out the spectator ions - SO₄^2-: Cu²⁺_(aq) + Zn_(s) → Zn²⁺_(aq) + Cu_(s)

 

[Haku]

thanks for the fast reply

so do i just need to put this equation Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s), thats what i put but thought that i need to incorporate further equations.

or would i need this as well CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

 

[Dreamerish*~]

thanks for the fast reply

so do i just need to put this equation Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s), thats what i put but thought that i need to incorporate further equations.

or would i need this as well CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

I think you'll be fine with the neutral species equation and half-equations.

 

[Haku]

thanks u rock

 

[paper cup]

I think you'll be fine with the neutral species equation and half-equations.

either or both?

 

[Dreamerish*~]

either or both?

For redox calculations, half-equations are a must.

I always chuck in a neutral species too, just to play it safe.

 

[Haku]

i guess it really depends on the marks allocated and the number of lines.