Question

Please do both and show all work! I need to compare the differences on how to do both.

1. Balance the reaction between Cd and Cr2O7^2- to form Cd^2+ and Cr^3+ in acidic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown.

___ Cd + ___ Cr2O7^2- ------> ___ Cd^2+ + ___ Cr^3+

Water appears in the balanced equation as a ______ (reactant, product, neither) with a coefficient of _____ (enter 0 for neither).

How many electrons are transferred in this reaction? ____

2. Balance the reaction between Br2 and Cl^- to form ClO^- and Br^- in basic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown.

___ Br2 + ___ Cl^- ------> ___ ClO^- + ___ Br^-

Water appears in the balanced equation as a ________ (reactant, product, neither) with a coefficient of ____ (enter 0 for neither).

How many electrons are transferred in this reaction? _____

Answer 1

1.

The unbalanced reaction is

Cd is being oxidised to Cd²⁺ . Hence, the oxidation half reaction is

We can balance the oxidation half reaction by simply adding 2e on the right to balance the charges.

Hence, the balanced oxidation half reaction is

Also, is being reduced to Cr³⁺ . Hence, the reduction half reaction is

Now, to balance the number of Cr atoms on both side we multiply 2 on Cr³⁺

Now, to balance the number of O atoms on both sides, we add 7H₂O on right.

Now, there are 14 H on the right and no H on left. Hence, we add 14 H⁺ on left to balance the number of H atoms.

Now, there are 14-2 = +12 charges on left and +6 charges on right. Hence, we will add 6 e on left to balance the charges.

Hence, the balanced reduction half reaction is

Now, there are 6e involved in reduction reaction and 2e involved in oxidation half reaction. Hence, to cancel the number of electrons, we multiply 3 throughout the oxidation half and add with the reduction half reaction to get the balanced redox reaction

+

=

Hence, the balanced reaction with the required coefficients is

Note: The required coefficients are shown in red.

Water appears in the balanced equation as a product with a coefficient of 7.

Note that each mole of Cd loses 2 moles of electron. There are 3 moles of Cd in the balanced equation.

Hence, total of 6 electrons are transferred in the reaction.

Hence, number of electrons transferred in the reaction = 6.

2.

The unbalanced reaction is

Br₂ is being reduced to Br^- . Hence, the reduction half reaction is

To balance the number of Br-, we ,ultiply 2 on Br^- .

Now, to balance the charges, we add 2e on left. Hence, the balanced reduction half reaction is

Now ,Cl^- is being oxidised to ClO^- . Hence, the oxidation half reaction is

To balance the number of O atoms, we add H2O on left.

Now, to balance the number of H atoms, we add 2 H+ on right

There are -1 charge on lef and +1 charge on right. Hence, we add 2 e on right to balance the charges.

Hence, the balanced oxidation half reaction is

Note that both half reactions involve 2 electrons. Hence, we can add them up and cancel the electrons to get the balanced reaction.

+

=

Hence, the balanced redox reaction with smallest integer coefficients as needed is

Water appears in the balanced equation as a reactant with a coefficient of 1.

Note that each moelcule of Br₂ loses 2e in the reduction reaction. There is only one molecule of Br2 in the balanced redox reaction.

Hence, the number of electrons transferred in this reaction is 2.