Question

consider the following unbalanced redox reaction Cr2O2−7(aq)+Cu(s)→Cr3+(aq)+Cu2+(aq)

a)Balance the equation: αCr2O2−7(aq)+βH+(aq)+γCu(s)→δCr3+(aq)+ϵH2O(l)+κCu2+(aq)

b)Determine the volume of a 0.950 M K2Cr2O7 solution required to completely react with 5.40 g of Cu.

Answer 1

(a) Balanced Chemial Equation:

Cr₂O₇^2- (aq) + 14 H⁺ + 3Cu 2Cr³⁺ + 3 Cu²⁺ + 7H₂O

That means = 1, = 14, = 3, = 2, = 7, = 3

It is a redox reaction. First calculate the oxidation state of the species included in the reaction. Oxidation state of
Cr in Cr₂O₇^2- is +6. Cu is in zero oxidation state on reactant side. Therefore now student can write two separate reaction.

Reduction half reaction: Cr₂O₇^2- (aq) Cr³⁺    (Gain of electrons)

Oxidation half reaction: Cu (s) Cu²⁺ (aq) (loss of electrons)

Now balance the coefficients in both reactions as follow:

Cr₂O₇^2- (aq) 2Cr³⁺

Cu (s) Cu²⁺ (aq)

Next balance the oxygen by using H₂O

Cr₂O₇^2- (aq) 2Cr³⁺ + 7H₂O

Next balance H with the help of H⁺

Cr₂O₇^2- + 14 H⁺(aq) 2Cr³⁺ + 7H₂O

Next balance positive charges with the help of e-

Cr₂O₇^2- + 14 H⁺(aq) + 6e- 2Cr³⁺ + 7H₂O

Cu (s) Cu²⁺ (aq) + 2e-

Now to cancel electrons multiply both equation with a whole number. In this case, mutiply first equation with 1st and 2nd equation with 3 and then add both equation to get final balanced equation.

Cr₂O₇^2- + 14 H⁺(aq) + 6e- 2Cr³⁺ + 7H₂O    

3Cu (s) 3Cu²⁺ (aq) + 6e-   

On adding the above two equations 6 electrons will cancel out on both sides and we will get final equation.

Cr₂O₇^2- (aq) + 14 H⁺ + 3Cu 2Cr³⁺ + 3 Cu²⁺ + 7H₂O

(b) First convert g of Cu into moles

No. of moles = m/M   (m = mass given in grams and M =molecular mass of Cu)

                  = 5.40/63.54 = 0.0849 moles

From the balanced chemical equation student can conclude that

1 mole of K₂Cr₂O₇ = 3 moles of Cu

Moles of K₂Cr₂O₇ = 0.0849/ 3 = 0.0283

Molarity = no of moles / Volume of solution (in L)

Put the value of molarity and number of moles in the formula and calculate the volume

0.950 = 0.0283 / Volume

Volume = 0.0283/ 0.950 = 0.0297L = 29.7 mL

Therefore volume needed comes out to be = 29.7 mL