202 Homework 9 8 10-REDOX & Electrochemistry (2017) 5 pts) Write the oxidation s

Question

202 Homework 9 8 10-REDOX & Electrochemistry (2017) 5 pts) Write the oxidation state for the underlined element in the box following each compou a) NaCIO2 d) HNO3 b) KPO4 c) H20 e) NaCN 5 pts) a) A Ph(NOs)2 solution containing a lead electrode is connected by means of a semi- perm membrane to a MgCl2 solut on containing a magnesium electrode. Which of the following cells representation of the spontaneous reaction that should result to generate electricity (c a) b) c) d) flow of flow of o flow of electrons fow of +2 CI NO No CI Pb Pb Mg CI Pb+2 Pb+ CATHODE ANODE Anic ANODE -+- CATHODE CATHODE- ANODE ANODE Anion fiow Anion flow Anion flow b) Briefly and clearly discuss your reasoning for the cell you selected from part a) below cell potential showing the two half-cell reactions and potentials.

Explanation / Answer

a)

Oxidation number of Na = +1

Oxidation number of O = -2

lets the oxidation number of Cl be x

we have below equation to be used:

1* oxidation number (Na) + 2* oxidation number (O) + 1* oxidation number (Cl) = net charge

1*(+1)+2*(-2)+1* x = 0

-3 + 1 * x = 0

x = 3

So oxidation number of Cl = +3

Answer: +3

b)

Oxidation number of K = +1

Oxidation number of O = -2

lets the oxidation number of P be x

we have below equation to be used:

3* oxidation number (K) + 4* oxidation number (O) + 1* oxidation number (P) = net charge

3*(+1)+4*(-2)+1* x = 0

-5 + 1 * x = 0

x = 5

So oxidation number of P = +5

Answer: +5

c)

Oxidation number of H = +1

lets the oxidation number of O be x

we have below equation to be used:

2* oxidation number (H) + 2* oxidation number (O) = net charge

2*(+1)+2* x = 0

2 + 2 * x = 0

x = -1

So oxidation number of O = -1

Answer: -1

d)

Oxidation number of H = +1

Oxidation number of O = -2

lets the oxidation number of N be x

we have below equation to be used:

1* oxidation number (H) + 3* oxidation number (O) + 1* oxidation number (N) = net charge

1*(+1)+3*(-2)+1* x = 0

-5 + 1 * x = 0

x = 5

So oxidation number of N = +5

Answer: +5

e)

Oxidation number of Na = +1

Oxidation number of N = +3

lets the oxidation number of C be x

we have below equation to be used:

1* oxidation number (Na) + 1* oxidation number (N) + 1* oxidation number (C) = net charge

1*(+1)+1*(+3)+1* x = 0

4 + 1 * x = 0

x = -4

So oxidation number of C = -4

Answer: -4

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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