Computations of a cell or solution potentials

[Dhamnekar Win,odd]

 

My  answers (a) 

The balanced reaction equation is 2VO₂⁺  + 3Zn  + H₂O ⇌  3ZnO + 2VO + 2H⁺

[math]E^{\circ}_{rxn} = E^{\circ}_{VO_2^+} - E^{\circ}_{Zn} = 1.00 V - (-0.76) V = 1.76 V [/math]

 

(b) [math] E = E^{\circ}_{Cu} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}[/math]

[math] E= 0.34 V - \frac{0.059}{2}\log \frac{1}{[0.10]}= 0.31 V [/math]

 

 I am working on other  questions.

Any chemistry help will be accepted.

[exchemist]

11 minutes ago, Dhamnekar Win,odd said:

 

My  answers (a) 

The balanced reaction equation is 2VO₂⁺  + 3Zn  + H₂O ⇌  3ZnO + 2VO + 2H⁺

E∘rxn=E∘VO+2−E∘Zn=1.00V−(−0.76)V=1.76V

 

(b) E=E∘Cu−0.059nlog1[Cu2+]

E=0.34V−0.0592log1[0.10]=0.31V

 

 I am working on other  questions.

Any chemistry help will be accepted.

Jolly good.

I'm not sure we ought to be offering to check all your answers, quite apart from the effort it requires (at least speaking for myself). If you've got specific queries, we can try to help with those.

 

 

[Dhamnekar Win,odd]

 I want to correct my answer to (a) The balanced reaction  is 2VO₂⁺  + Zn⁰  + 2H⁺  ⇌  2VO²⁺   +  Zn²⁺   + H₂O

 

Answer to (c) Reduction: [math]E = E^{\circ}_{Ag} - \frac{0.059}{n}\log{ \frac{1}{[Ag^+]}}= 0.80 V - \frac{0.059}{1} \log{\frac{1}{[0.035]}} =  0.71 V[/math]

Oxidation: [math]E= E^{\circ}_{Fe} - \frac{0.059}{n} \log {\frac{[Fe^{2+}]}{[Fe^{3+}]}}= 0.77 V - \frac{0.059}{1} \log{\frac{[0.0010]}{[0.010]}} = 0.83 V[/math]

E_cell = E_red  - E_ox = 0.71 - 0.83 = -0.12 V  

[Dhamnekar Win,odd]

I am doubtful about my answer to (d).

My attempt to answer (d):

Reduction: E = E°_AgCl  - [math]\frac{0.059}{n} \log {[Cl^-]} = 0.22 V  - \frac{0.059}{1} \log{[0.20]} = 0.26 V[/math] 

Oxidation: E= E°_TiO²⁺  - [math]\frac{0.059}{n} \log {\frac{[Ti^{3+}]}{[Ti^{2+}] [H^+]^2}} [/math] =[math] 0.10 V - \frac{0.059}{1}\log {\frac{[0.010]} {[0.20][0.0020]^2}}= -0.14 V [/math]

E_cell   = E_red  - E_ox = 0.26 V - (-0.14 V) = 0.40 V Is this answer correct?

Edited May 3, 2022 by Dhamnekar Win,odd

[NTuft]

20 hours ago, Dhamnekar Win,odd said:

I want to correct my answer to (a) The balanced reaction  is 2[[VO₂]=4O]⁺  + Zn⁰  + 2H⁺  ⇌  2VO[2O]²⁺   +  Zn²⁺   + H₂O[3O]

 

23 hours ago, Dhamnekar Win,odd said:

The balanced reaction equation is 2[[VO₂=4O]]⁺  + 3Zn  + H₂O[5O] ⇌  3ZnO[3O] + 2VO[} + 2H⁺

Edited May 3, 2022 by NTuft

[studiot]

43 minutes ago, NTuft said:

 

I've not seen that method of mass accounting before, well done for spotting the issue and presenting it.  +1

However perhaps you should explain the meaning of your square brackets, some may have difficulty working it out for themselves.

[NTuft]

A standard technique in editorialization is to insert [Ed: ]

How can you operate latex on this site, or where would you recommend I find it, please.

Edited May 3, 2022 by NTuft
coding

[swansont]

5 minutes ago, NTuft said:

A standard technique in editorialization is to insert [Ed: ]

How can you operate latex on this site, or where would you recommend I find it, please.

https://www.scienceforums.net/topic/116044-math-test/

Scroll down for the latest info

[Dhamnekar Win,odd]

7 hours ago, NTuft said:

 

  Hi, 

   Thanks for finding out the fault in  my answer.

  Now, I rewrite the corrected balanced  reaction 2VO₂⁺  + Zn⁰  + 4H⁺  ⇌ 2VO²⁺  + Zn²⁺  + 2H₂O 

[Dhamnekar Win,odd]

12 hours ago, Dhamnekar Win,odd said:

  Hi, 

   Thanks for finding out the fault in  my answer.

  Now, I rewrite the corrected balanced  reaction 2VO₂⁺  + Zn⁰  + 4H⁺  ⇌ 2VO²⁺  + Zn²⁺  + 2H₂O 

If I write the balanced equation as 2VO₂⁺  + Zn⁰  + 2H⁺ ⇌ 2VO²⁺  + Zn²⁺O^-2 + H₂O

Is this correct? 

[Dhamnekar Win,odd]

Answer to question(d):

                                                      Co³⁺                         +                 V²⁺                         →                      Co²⁺                          +                      V³⁺

 

 

Defined redox couple is  Co³⁺/Co²⁺ system

[math] [Co^{2+}] = \frac{0.0050 mol}{0.200L} =0.025 M, [Co^{3+}]= \frac{0.010 mol}{0.200 L}=0.050 M, E_{sol}= E^{\circ}_{Co} -\frac{0.059}{n} \log{\frac{[Co^{2+}]}{[Co^{3+}]}} = 1.82 V - \frac{0.059}{1}\log{\frac{[0.025]}{[0.050]}} =1.84 V [/math]

 

 

Edited May 4, 2022 by Dhamnekar Win,odd

[NTuft]

On 5/3/2022 at 10:12 PM, Dhamnekar Win,odd said:

On 5/3/2022 at 10:04 AM, Dhamnekar Win,odd said:

Now, I rewrite the corrected balanced  reaction 2VO₂⁺  + Zn⁰  + 4H⁺  ⇌ 2VO²⁺  + Zn²⁺  + 2H₂O 

If I write the balanced equation as 2VO₂⁺  + Zn⁰  + 2H⁺ ⇌ 2VO²⁺  + Zn²⁺O^-2 + H₂O

Is this correct? 

I do not know.

Perhaps split apart the reduction from the oxidation reaction.

What is oxidized, what is reduced? Is the start => change  of the solution set-up to allow this reaction? 

Remember they want the E°.

Edited May 6, 2022 by NTuft