I) Do the same as in the previous question (also see the NOTE). Fe(s) + Cu2*(aq) Fe3+(aq) + Cu(s) ----> (Place ONLY #s in the spaces when balancing) OA = : RA = II) Give the oxidation number (ON) of S in the following: a) SOCI2 : ON = b) H2S2: ON = c) H2SO4: ON = d) Na2s: ON =

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Chapter1: Chemical Foundations
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**QUESTION 14**

I) Do the same as in the previous question (also see the **NOTE**).

\[ \_\_\_ \, \text{Fe(s)} + \_\_\_ \, \text{Cu}^{2+}\text{(aq)} \rightarrow \_\_\_ \, \text{Fe}^{3+}\text{(aq)} + \_\_\_ \, \text{Cu(s)} \]

(*Place ONLY \#s in the spaces when balancing*)

**OA** = \_\_\_\_\_ ;  **RA** = \_\_\_\_\_ 

II) Give the oxidation number (ON) of S in the following:

a) \( \text{SOCl}_2 \) : **ON** = \_\_\_\_\_

b) \( \text{H}_2\text{S}_2 \) : **ON** = \_\_\_\_\_

c) \( \text{H}_2\text{SO}_4 \) : **ON** = \_\_\_\_\_

d) \( \text{Na}_2\text{S} \) : **ON** = \_\_\_\_\_
Transcribed Image Text:**QUESTION 14** I) Do the same as in the previous question (also see the **NOTE**). \[ \_\_\_ \, \text{Fe(s)} + \_\_\_ \, \text{Cu}^{2+}\text{(aq)} \rightarrow \_\_\_ \, \text{Fe}^{3+}\text{(aq)} + \_\_\_ \, \text{Cu(s)} \] (*Place ONLY \#s in the spaces when balancing*) **OA** = \_\_\_\_\_ ; **RA** = \_\_\_\_\_ II) Give the oxidation number (ON) of S in the following: a) \( \text{SOCl}_2 \) : **ON** = \_\_\_\_\_ b) \( \text{H}_2\text{S}_2 \) : **ON** = \_\_\_\_\_ c) \( \text{H}_2\text{SO}_4 \) : **ON** = \_\_\_\_\_ d) \( \text{Na}_2\text{S} \) : **ON** = \_\_\_\_\_
**Balance each equation then identify the oxidizing agent (OA), the reducing agent (RA), the species oxidized, and the species reduced:**

**{NOTE:**
i)  Put only numbers in the spaces to balance the equations even if it is the number “1”.
ii)  For the OA, RA, species oxidized, and species reduced, **write the complete formula of the substance WITHOUT THE PHASE.** (e.g., if H in HCl gets reduced then the OA should be written as HCl; if the P in P₄ gets oxidized then the RA should be written as P₄.
iii)  **Do NOT use subscripts or superscripts in formula answers.** Put numbers on the same level as symbols (e.g. “N₂” would be written as N2 in answer.)}

a)  
\[ \_\_\_ \text{S}_8(s) + \_\_\_ \text{F}_2(g) \rightarrow \_\_\_ \text{SF}_4(g) \]
Species oxidized = \[ \_\_\_ \]

**(Place ONLY #s in the spaces when balancing)**

b)  
\[ \_\_\_ \text{CaI}_2(aq) + \_\_\_ \text{Li}(s) \rightarrow \_\_\_ \text{LiI}(aq) + \_\_\_ \text{Ca}(s) \]

**(Place ONLY #s in the spaces when balancing)**

OA = \[ \_\_\_ \] ; RA = \[ \_\_\_ \]
Transcribed Image Text:**Balance each equation then identify the oxidizing agent (OA), the reducing agent (RA), the species oxidized, and the species reduced:** **{NOTE:** i) Put only numbers in the spaces to balance the equations even if it is the number “1”. ii) For the OA, RA, species oxidized, and species reduced, **write the complete formula of the substance WITHOUT THE PHASE.** (e.g., if H in HCl gets reduced then the OA should be written as HCl; if the P in P₄ gets oxidized then the RA should be written as P₄. iii) **Do NOT use subscripts or superscripts in formula answers.** Put numbers on the same level as symbols (e.g. “N₂” would be written as N2 in answer.)} a) \[ \_\_\_ \text{S}_8(s) + \_\_\_ \text{F}_2(g) \rightarrow \_\_\_ \text{SF}_4(g) \] Species oxidized = \[ \_\_\_ \] **(Place ONLY #s in the spaces when balancing)** b) \[ \_\_\_ \text{CaI}_2(aq) + \_\_\_ \text{Li}(s) \rightarrow \_\_\_ \text{LiI}(aq) + \_\_\_ \text{Ca}(s) \] **(Place ONLY #s in the spaces when balancing)** OA = \[ \_\_\_ \] ; RA = \[ \_\_\_ \]
Expert Solution
Step 1

Oxidation number is defined as an number is assigned to an element in chemical combination which represents the number of electrons gained(if positive ) or looses(if negative ) by an atom of that element in compound 

Negative charge comes on that atom which is high electron negative atom and positive charge comes on that which is less electron negative atom 

Oxidising agent which gain electron 

Reducing agent which loose electron.

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