Chemistry Forum

Author's given answer is wrong.The correct answer is + 1.21 volts using the formula for ∆G°= -nFE°

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

Would you explain that?

I think your answer is correct. There is a typographical error in the given answer.

Author has given the following answer: is there typographical error in the above answer?

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°...

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?

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...

So, (x + y) is a stoichiometry coefficient. Isn't it?

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?

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?

I am working on a) and b).

(Deviation from the subject)What is PN? Would you explain?

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...

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...

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?

Description of 'AmmoniaConcentration' package in 'R'. The total ammonia in aqueous solution is present in two chemical species: un-ionized ammonia, NH3 , and the ionized form, NH4 +. The NH3 species is the one more toxic for aquatic organisms, but current analytical methods do not permit measurement...