Posted before but was given the wrong answer! Thank you! Need help on 3,4 and 5!
ID: 1000612 • Letter: P
Question
Posted before but was given the wrong answer! Thank you! Need help on 3,4 and 5! Br2, CuCl2, CHi, CO2, Br20, 2. Which of the following are redox reactions? a. 3H2(g) + N2(g) 2 NH3 (g) b, C(s) + O2(g) CO2 (g) c. II:CCH2 (g) + H2 (g) HCCI, (g) d. Cu2+ (aq) + 4 NH3 (aq) Cu(NH3)42+ (aq) e. 3Cu(s) + 8H+ (aq) + 2 NO' (aq) 3 Cu2+ (aq) + 2 NO(g) + 4 H2O (l) 3. Balance the following redox reaction: In base: F (a)+ So. (aa)- 1a 6)+s(6 3. Balance the following redox reaction: In base: 1. (aq) + sos (aq)12 (s) + S (s) Use the tabhl of standard reduction potcatiah for fhe f lw inies 4. Predict if the following will occur under standard conditions, if so, write the prod ucts wi correct phases: a) HCI (aq) + Au (s) Find a reagent that can oxidize Br' to Br2 but cannot oxidize CI1 to Cl2 b) HCI (aq) + Na (s) c) Au (aq)+Na (s) 5. 6. Assuming standard conditions, indicate which of the following is true: a. H2 (g) can reduce Ag' (aq) b. H2 (g) can reduce Ni2 (aq) c. Fe (aq) can reduce Cu'(aq) d. H (aq) can oxidize Mg(s). e. Pb2 (aq).can oxidize Ni(s). 2+ 7. You decide to construct a zinc/aluminum galvanic cell in which the electrodes are co by a wire and the solutions are connected with a salt bridge. One electrode consists o aluminum bar in a 1.0 M solution of aluminum nitrate. The other electrode consists bar in a 1.0 M solution of zinc nitrate. a. Which electrode is the cathode and which is the anode? b. What is the direction of electron flow? c. What chemical reactions are occurring at each electrode? d. What is the overall balanced chemical reaction? e. After a period of time, will the bar of zinc become heavier, lighter or stay th What about the aluminum bar? in the standard cell potential?
Explanation / Answer
Answers to 2 and 3)
2)
Redox reaction is the combination of reduction and oxidation reaction.
In Redox reaction one species get reduced by anathor species.
The species reduced get lowered in oxidation number (O.N.). and the one which get oxidized increases in Oxidation number.
In elemental or homoatomic state O.N. is 0.
O.N per atom shown.
a) H2 + N2 ---------> 2NH3
O.N. 0 0 -3 +1
N2 reduced from 0 (in N2) -------> -3 (in NH3)
H2 oxidized from 0 (in H2) -- -------> +1 in (NH3)
b) C + O2 ---------> CO2
O.N. 0 0 +4 -2
c) H2C = CH2 + H2 -------> H3C-CH3
In H2C = CH2 net charge 0. O.N. of H in Carbon compound is +1 and hence
We write
4 x ON of H + 2 x ON of C = 0
4 x (+1) + 2 x ON of C = 0
2 x ON of C = –4
On of C = –2
Simillarly,
In H3C-CH3,
6 x ON of H + 2 ON of C = 0
ON of C = –3
It means
Here also reduction and oxidation occur.
In ON of H changes from 0 in H2 -------> +1 in H3CCH3
ON of C changes from –2 in H2C=CH2 -------------> –3 in H3C-CH3
d) Cu2+ (aq.) + NH3 -------à [Cu(NH3)4]2+ (aq.)
NH3 is neutral ligand and Cu2+ remains Cu2+ in complex
Hence it’s not a redox reaction.
e) 3Cu (o) + 8H+ (aq.) + 2NO3–. -------> Cu2+ (aq.) + 2NO + 4H2O
ON 0 +1 +5 (for N) +2 for Cu2+ +2 for N
It’s a redox reaction as,
Cu oxidizes from 0 in Cu (s) ---------> +2 in Cu2+ ………….(ON increases)
N reduced from +5 in NO3 – -----------> +2 in NO …………(ON decreases)
Hence Redox reactions are (a), (b), (c), (e).
(d) is not redox reaction.
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3) Given unbalanced redox reaction,
I–(aq.) + SO42– (aq.) -----------------> I2 (s) + S (s)
Let us write and balance half-cell reactions
Oxidation Half Reaction (OHR): I–(aq.) --------------> I2 (s)
Reduction Half reaction (RHR) : SO42– (aq.) -----------------> S (s)
Balancing of half cell reactions,
1) Balancing of atoms other than O and H
OHR: 2 I–(aq.) -----------------> I2 (s)
RHR: no need they are balanced already.
2) Balancing of O and H by adding H2O and H+ ions on appropriate sides,
OHR : No need
RHR: SO42– (aq.) + 8H+ (aq.) -----------------> S (s) + 4H2O
3) Balancing of charge by adding electrons on appropriate side.
OHR: 2 I–(aq.) -----------------> I2 (s) + 2 e–
RHR: SO42– (aq.) + 8H+ (aq.) + 6 e– -----------------> S (s) + 4H2O
4) Balancing of electron number in both halves,
We need to just multiply OHR by 3 so electron number will be 6.
Hence,
OHR: 6 I–(aq.) -----------------> 3I2 (s) + 6 e–
RHR: SO42– (aq.) + 8H+ (aq.) + 6 e– -----------------> S (s) + 4H2O
Let us two Half reactions,
6 e – common to both sides get cancelled out.
6 I–(aq.) + SO42– (aq.) + 8H+ (aq.) -----------------> 3I2 (s) + S (s) + 4H2O
Hence the balance redox reaction in acidic medium.
6 I–(aq.) + SO42– (aq.) + 8H+ (aq.) -----------------> 3I2 (s) + S (s) + 4H2O.
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