Under certain conditions, H_2O_2 can act as an oxidizing agent; under other cond

Question

Under certain conditions, H_2O_2 can act as an oxidizing agent; under other conditions, as a reducing agent. What is the best theoretical explanation for this? (A) H_2O_2 is a good bleaching agent. (B) Peroxides are stronger oxidizing agents than are oxides. H_2O_2 will decolorize KMnO_4 solutions in the presence of an acid and will turn black lead sulfide to a white compound. An atom within a compound can sometimes attain a more stable electronic structure either by gaining or by losing electrons. ____ Which of these changes will produce the most positive voltage for this half reaction in the direction written? Co^2+ + 2e^- rightarrow Co() E degree = -0.28 VV (A) increasing the amount of solid Co (B) decreasing the amount of solid Co (C) increasing the concentration of Co^2+ (D) decreasing (be concentration of Co^2+ ____ In which case would the least number of faradays of electricity be required for the liberation of 1.0 g of free metal? Atomic Molar Masses k 39 1 g-mol^-1 Na 23.0 g-mol^-1 Cu 63.5 g-mol^-1 Ag 107.9 g-mol^-1 (A) K from molten KOH (B) Ni from molten NaCl (C) Cu from aqueous CuSO_4 (D) Ag from aqueous AgNO_3 A vanadium electrode is oxidized electrically, If the mass of the electrode decreases by 114 mg during the passage of 650 coulombs, what is the oxidation state of the vanadium product? (A) +1 (B) +2 (C) +3 (D) +4 A current of 5.00 A is passed through an aqueous solution of chromium (III) nitrate for 30.0 min. How many grams of chromium metal will be deposited at the cathode? (A) 0.027 g (B) 1.62 g (C) 4.85 g (D) 6.33 g ___What mass of platinum could be plated on an electrode from the electrolysis of a Pt(NO_3)_2 solution with a current of 0.500 A for 55.0 s? (A) 27.8 mg (B) 45.5 mg (C) 53.6 mg (D) 91.0 mg

Explanation / Answer

(25) (D)

(26)

According to Nernst equation of electrode potential,

Eele = E0ele - (0.0591/n) *Log1/[Co2+]

So, to get more positive electrode potential value, [Co2+] should be increased.

(C)

(27)

(A) To produce 39.1 g. 1 F is needed

for 1g. of K = 1/39.1 = 0.0256 F is neede

(B) For 23.0 g. of Na 1 F is needed

then for 1.0 g. of Na 1 / 23 = 0.0435 F is needed

(C) To produce 63.5 g. of Cu 2 F is needed

then, for 1 g. of Cu = 2 / 63.5 = 0.0315 F is needed

(D) To produce 107.9 g, of Ag 1 F is needed

then for 1.0 g. of Ag = 1/107.9 = 0.00927 F is needed

Hence the answer is (D)

(29) According first law of electrolysis,

W = M c t / Z F
W = 52 * 5.00 * 30.0 * 60 / (3 * 96500)

W = 1.61 g.

(B)

(30)

W = 195 * 0.500 * 55.0 / (2 * 96500)

W = 0.0278 g. = 27.8 mg

(A)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.