Question

Tro, Nivaldo 4th ed, Introductory Chemistry Chapter 16, Exercise 98 Please complete the math and solve Exercise 98:

98. Part a: Consider the unbalanced redox reaction:Cr₂ O₇^2-(aq) + Cu(s) →Cr3+(aq) + Cu²⁺

Balance the equation in acidic solution

Part b: and determine how much of a 0.850 M K₂ Cr₂ O₇ solution is required to completely dissolve 5.25 g of Cu.

Answer 1

Cr2 O72-(aq) + Cu(s) →Cr3+(aq) + Cu2+

Split in red and ox reactions

Cr2O7-2 --> Cr+3

Cu -> Cu2+

balance Chromium

Cr2O7-2 --> 2Cr+3

Cu -> Cu2+

Balance O

Cr2O7-2 --> 2Cr+3 + 7H2O

Cu -> Cu2+

Balance H

14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O

Cu -> Cu2+

Balance charges

14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O

(14+ -2 = +12 in the left side; 2*3 = +6 in the right side... add 6 electrons in the left)

6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O

Cu -> Cu2+ + 2e-

Balance electrons

6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O

3Cu -> 3Cu2+ + 6e-

add both equatins

3Cu + 6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O + 3Cu2+ + 6e-

Cancel common terms

3Cu + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O + 3Cu2+

b)

how much is needed M = 0.85 of K2CrO7 to dissolve 5.25 g of Cu

mol of Cu = masS/MW

MW = 63.5 g/mol

mol Cu = 5.25/63.5 = 0.08267 mol of Cu

then, acorrding to the ratio

1 mol of Cr2O7-2 will dissolve 3 mol of Cu

ratio is 1:3 or 1/3

therefore

mol of Cr2O7-2 needed

0.08267/3 = 0.02755 mol of Cr2O7-2 needed

therefore, since ratio is 1:1 with respect to K2Cr2O7 : CrO7-2

then we need 0.02755 mol of K2Cr2O7

how much volume needed:

M = mol/V

V = mol/M = 0.02755/0.85 = 0.03241 L

V = 3.24 ml