Select only option Fall 2017 8. Balance the following redox reaction in acidic s

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Fall 2017 8. Balance the following redox reaction in acidic solution: NO3-(aq) + Cu(s) NO(g) + Cu"(aq) What is the SUM of all the coefficients once balanced? a. 10 b. 12 c. 15 d. 22 e. 25 9. Which of the following reactions is an oxidation-reduction reaction? Mg(OH)2(s) Mg2+(aq) + 20H(aq) NH,(g) + H2O(l)NH4OH(aq) 2H20(I) H,O"(aq) +OH(aq) CO(g) + H2O(g) CO2(g) + H2(g) b. c. d. 10. What is the reducing agent in the following reaction below? 2N02(g) + 7H2(g) 2NH3(g) + 4H20(g) a. NO2 b. NH3 c. H20 d. H2

Explanation / Answer

Answer:

Given unbalanced redox reaction:

NO3- (aq) + Cu (s) ----------> NO (g) + Cu2+ (aq)

Let's first write Half-cell reactions,

Oxidation Half Reaction (OHR) : Cu (s) ----------> Cu2+ (aq)

Reduction Half Reaction (RHR) : NO3- (aq) ----------> NO (g)

We will balance these 2 half cell reaction separately and then add them.

Step-1) Balancing of atoms other than H & O.

They are already balanced. So step is skipped.

Step-2) Balancing of O & H.

To balance O we add the appropriate number of H2O molecules on O deficient side to balance.

OHR: Not necessary

RHR: NO3- (aq) ----------> NO (g) + 2 H2O

Then to balance H we add excess number of H+ ions on opposite side so we have,

RHR: NO3- (aq) + 4 H+ (aq) ----------> NO (g) + 2 H2O.

Step-3) Balancing of charge: We add the appropriate number of electrons on the side of excess +ve charge.

OHR: Cu (s) ----------> Cu2+ (aq) + 2e-.

RHR: NO3- (aq) + 4 H+ (aq) + 3e- ----------> NO (g) + 2 H2O.

Step-4) Balancing of electron number in both half-cell reaction. On observation one can figure out that OHR need to be multiplied by 3 and RHR need to be multiplied by 2 (6 is LCM of 2&3).

SO we have

OHR: 3 Cu (s) ----------> 3 Cu2+ (aq) + 6e-.

RHR: 2 NO3- (aq) + 8 H+ (aq) + 6e- ----------> 2NO (g) + 4 H2O.

Step-5) Adding 2 Half-cell reactions,

6e- being same cancelled out mutually from both sides.So,

2 NO3- (aq) + 3 Cu (s) + 8 H+ (aq) + ----------> 2NO (g) + 3 Cu2+ (aq) + 4 H2O

is the balanced Redox equation in acidic medium.

Sum of the coefficients = 2 + 3 + 8 + 2 + 3 + 4 = 22

Answer option (d) 22.

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