8. (5 pts) Write the oxidation state for the underlined element in the box follo

Question

8. (5 pts) Write the oxidation state for the underlined element in the box following each compound. a) NaCIO.2 b) K3PO4 c) H202 d) HNO3 e) NaCN 9. (5 pts) a) A Pb(NO3)2 solution containing a lead electrode is connected by means of a semi-permeable membrane to a MgCl2 solution containing a magnesium electrode. Which of the following cells is a correct representation of the spontaneous reaction that should result to generate electricity (circle your answer) a) b) d) flow of electrons flow of flow of flow of NO C CI NO3 Pb2 N Pb Mg Mg Pb Pb +2 Pb CI ANODECATHODE Anion flow CATHODEANODE Anion flow ANODE CATHODE ANODE 4-CATHODE Anion flow Anion flow b) Briefly and clearly discuss your reasoning for the cell you selected from part a) below and calculate the cell potential showing the two half-cell reactions and potentials.

Explanation / Answer

8)

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

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

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

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

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

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