## Search found 243 matches

Sat Aug 20, 2022 9:18 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

The table in your link says Ti³⁺ + 3 e⁻ ⇌ Ti⁰ E° = -1.208 V

Reverse it for oxidation of Titanium metal Ti⁰ ⇌ Ti³⁺ + 3 e⁻ E° = +1.208 V
Sat Aug 20, 2022 7:49 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

Author's given answer is wrong.The correct answer is + 1.21 volts using the formula for ∆G°= -nFE°
Sat Aug 20, 2022 4:18 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

Why did you choose ① as reduction half reaction and ② as oxidation half reaction?

Would you explain that?
Sat Aug 20, 2022 4:02 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

Sat Aug 20, 2022 3:57 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### Re: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

Author has given the following answer:
E⁰ for Ti.png (7.9 KiB) Viewed 502 times
is there typographical error in the above answer?
Sat Aug 20, 2022 2:17 am
Forum: Chemistry forum
Topic: E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻
Replies: 9
Views: 512

### E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

It is sometimes necessary to compute E° for a given half reaction from other half reaction of known E°. For example, the standard electrode potential for the oxidation of Titanium metal to Ti³⁺ can be obtained from the following half reaction: Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts ① Ti³⁺ + e⁻ ⇌ Ti²⁺ E°...
Thu Aug 18, 2022 10:58 pm
Forum: Chemistry forum
Topic: Solubility product constant
Replies: 8
Views: 5009

### Re: Solubility product constant

I found on Internet, E°= 0.37 Volts for the reaction [Ag(NH₃)₂]⁺ + e⁻ ⇌ Ag + 2 NH₃.

So, K = Kf, Log Kf= $$\frac{0.80 Volts-0.37 Volts}{0.059} = 7.29 K_f= 1.9 \times 10^7$$

There is a discrepancy between your answer of 2.1e6 and my answer 1.9e7. Is this discrepancy significant?
Thu Aug 18, 2022 9:06 am
Forum: Chemistry forum
Topic: Solubility product constant
Replies: 8
Views: 5009

### Re: Solubility product constant

My answer to (a): Mg(OH)_2 + 2 e^- \leftrightarrow Mg + 2 OH^- E^{\circ} = -2.67 V oxidation Mg + 2 e^- \rightarrow Mg E^{\circ}= -2.36 V K = \frac{1}{K_{sp}} reduction log K = \frac{2 [-2.36 -(-2.67)]}{0.059} =10.51 K=\frac{1}{K_{sp}}= 3.2 \times 10^{10} K_{sp} = 3.1e-11 My answer to (b): Ag + 2 N...
Wed Aug 17, 2022 10:25 pm
Forum: Chemistry forum
Topic: Dissolution of alkali metals in liquid ammonia
Replies: 3
Views: 357

### Re: Dissolution of alkali metals in liquid ammonia

So, (x + y) is a stoichiometry coefficient. Isn't it?
Wed Aug 17, 2022 10:31 am
Forum: Chemistry forum
Topic: Dissolution of alkali metals in liquid ammonia
Replies: 3
Views: 357

### Dissolution of alkali metals in liquid ammonia

The alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature.

$$M + (x + y ) NH_3 \rightarrow [ M (NH_3)_x]^+ + [ e (NH_3)_y ]^-$$

What is x and y in the above equation?

How to interpret the above chemical equation?
Wed Aug 17, 2022 2:41 am
Forum: Chemistry forum
Topic: Na2O2 + CH3NO2 =?????
Replies: 1
Views: 232

### Na2O2 + CH3NO2 =?????

I know reactants and products of the following chemical reaction

$$3Na_2O_2 + CH_3NO_2 + 2H_2O \leftrightarrow HNO_3 + CO + 6NaOH$$

I want to know the product of the following reaction.

$$Na_2O_2 + CH_3NO_2 =???$$

Would any member of chemistry forum answer my question correctly?
Wed May 25, 2022 11:49 pm
Forum: Chemistry forum
Topic: Solubility product constant
Replies: 8
Views: 5009

### Re: Solubility product constant

I am working on a) and b).

(Deviation from the subject)What is PN? Would you explain?
Wed May 25, 2022 9:43 pm
Forum: Chemistry forum
Topic: Solubility product constant
Replies: 8
Views: 5009

### Solubility product constant

a)Calculate the solubility product constant for Mg(OH)2 Mg(OH)2 + 2 e⁻ → Mg + 2OH⁻ E° = -2.67 b) Calculate the formation constant for Ag(NH3)2⁺ How to answer both these questions? (Deviation from the subject) Why don't other moderators answer the questions asked by chemistry students, chemistry prof...
Fri May 06, 2022 9:17 pm
Forum: Chemistry forum
Topic: Normality question
Replies: 3
Views: 10101

### Re: Normality question

Though, your answer looks to me correct, my doubt is if 173.80 g is equivalent weight, how can we take it as a "Molar mass"? I think in this problem, we should assume (1equivalent acid/1 equivalent base) 20.07 mL x (1 L/1000 mL) x (0.1100 eq base/1 L) x (1 eq acid/1 eq base) x (173.8 g/1 e...
Fri May 06, 2022 4:36 am
Forum: Chemistry forum
Topic: Normality question
Replies: 3
Views: 10101

### Normality question

Find the percent acid (eq wt 173.80) if 20.07 mL of 0.1100 N base is required to neutralize 0.721 g of a sample.

What is the answer to this question?