The element Ba in the reactants has a charge of zero, but in the BaClh product,

Question

The element Ba in the reactants has a charge of zero, but in the BaClh product, t is present a becomes more positive. The element in Cly starts with a charge of zero in the reactants and becoming more negative. You can use the acronym OIL RIG, which stands for "Oxidation is Lo identily that Ba has been oxidized and Cl has been reduced over the course of the reaction Constants 1 Periodic Table In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced You may want to reference (Pages 238-240) Section 7.6 while completing this problem. Part B Br2(g) + 2K1(aq) 2KBr(aq) + 12(s) 0 1 (in KI) gains electrons and is oxidized. Br2 loses electrons and is reduced. O Br2 gains electrons and is oxidized. I (in KI) loses electrons and is reduced O I (in KI) loses electrons and is oxidized. Br2 gains electrons and is reduced O Br2 loses electors and is oudized.(in KI) gains electrons and is reduced. Submit Part C 2Al(s) + 3Sn2+ (aq)- 2AP+ (aq) + 3Sn(s) O Sn loses electrons and is oxikdized. Al gains electrons and is reduced O Al loses electrons and is oxidized. Sn2 . gains electrons and is reduced. O Sn2 gains electrons and is oxidized. Al loses electrons and is reduced O Al gains electrons and is oxidized. Sa2 loses electrons and is reduced Submit Part D Fe(s) + Pb(NOJ2(aq) Fe(NOJ2(aq) + Pb(s)

Explanation / Answer

Part B

Br2 + 2KI ----- 2KBr + I2(s)

Bromine is reducing from oxidation state of 0 to oxidation state of -1

Iodine is oxidizing from oxidation state of -1 to oxidation state of +1

Therefore, correct answer is Option C

I- in KI is losing electron and oxidizing wheres Br2 gain electrons and is reduced

Part C

2Al + 3Sn(2+) ------ 2Al(3+) + 3Sn(s)

Sn is reducing from oxidation state of 2 to oxidation state of 0

Al is oxidizing from oxidation state of 0 to oxidation state of +3

Therefore, correct answer is Option B

Al looses electrons and is oxidized. Sn(2+) gains electrons and is reduced

Part D

Fe(s) + Pb(NO3)2 ---- Fe(NO3)2 + Pb(s)

Pb is reducing from oxidation state of +2 to oxidation state of 0

Fe is oxidizing from oxidation state of 0 to oxidation state of +2

No options are provided. hence provided the oxidizing agent and reducing agent

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