Balancing Redox Reactions 41. Balance each redox reaction occurring in basic aqu

ID: 1065072 • Letter: B

Question

Balancing Redox Reactions 41. Balance each redox reaction occurring in basic aqueous solution. a. H202(aq) Clo2(aq) 37. Balance each redox reaction occurring in acidic aqueous solution b. Al(s) MnO4 (aa) Mno2 (s) Al(OH)4 (aq) a. KOs) Cr (aq) Cr(s) K+(aq) c. Cl20g) Cl (aa) CIO (aq) b. Al(s) Fe2 (aa) A13+(aq) Fe(s) 42. Balance each redox reaction occurring in basic aqueous solution. c. Bro3 (aa) N2H4(g) Br (aq) N20g) a. Mno4 (aa) Br (aa) Mno20s) BrO3 (aa) 38. Balance each redox reaction occurring in acidicaqueous solution. b. Ag (s) CN (aa) O20g) Ag(CN)2 (aa) a. Zn(s) Sn2 (aa) Zn2+ aq) Sn(s) c. NO2 (aa) Al(s) NH3(g) AlO2 (aq) Mg (aa) Cr(s) 3+ Voltaic cells, Standard Cell Potentials, and Direction 39 Balance each redox reaction occurring in acidic aqueous solution. of spontaneity a. PbO2 (s) I (aq) b. SO32 (aq) Mnoa (aq) SO4 (aq) Mn (aq) 2+ 43. Sketch a voltaic cell for each redox reaction. Label the anode c. SaO3 (aa) C120g) SO42 (aa) Cl (aq) and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate 40. Balance each redox reaction occurring in acidic aqueous solution. the direction of electron flow. a. I (aa) NO2 (aa) I2(s) NO(g) b. Cio4 (aa) Cl (aa) CIO3 (aa) Cl2(g)

Explanation / Answer

QUESTION 37

A corrosive is an atom or particle equipped for giving a hydron (proton or hydrogen particle H+), or, then again, fit for shaping a covalent bond with an electron combine .

A Lewis base, then, is any species that gives a couple of electrons to a Lewis corrosive to shape a Lewis adduct. For instance, OH and NH3 are Lewis bases, since they can give a solitary combine of electrons.

a. K(s)+Cr3+(aq)=Cr(s)+K+(aq)

Balance each half-reaction separately.

K ==> K+ . . .add one e- to the right side to balance the charge.
K ==> K+ + e-

Cr3+ ==> Cr . . .add 3e- to the left side to balance the charge.
Cr3+ + 3e- ==> Cr

Note that the K equation has one e- on the right and the Cr equation has 3e- on the left side. Multiply the K equation by 3 so that when we add it to the Cr equation, the electrons will cancel.

.3K ==> 3K+ + 3e-
+Cr3+ + 3e- ==> Cr

.3K + Cr3+ ==> 3K+ + Cr

3 K + Cr3+ = 3 K+ + Cr-----------------answer

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b. Al (s)+Fe2+(aq)=Al3+(aq)+Fe(s)

Al = Al3+ + 3e- ) x 2
Fe2+ + 2e- = Fe ) x 3

2 Al (s)+ 3 Fe2+(aq)+ = 2 Al3+(aq)+ + 3 Fe(s)

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c. Bro3 (aq)+ N242-(aq)=Br (aq)+N2(s)

2 BrO3(aq) + 3 N2H4(g) = 2 Br(aq) + 3 N2(g) + 6 H2O(l)

Reaction stoichiometry

Compound

Coefficient

Molar Mass

BrO3(aq)

127.9022

N2H4(g)

32.04516

Br(aq)

79.904

N2(g)

28.0134

H2O(l)

18.01528