Question

Suppose that you have to balance the following half-reaction in basic solution.

Cr₂O₇^2??(aq) ? Cr³⁺?(aq)

What is the coefficient on water of the final balanced half-cell reaction?

	A) 5
	B) 6
	C) 7
	D) 8

Answer 1

While balancing a half-cell reaction in basic medium we need to go through the following steps

step:1:Write down the unbalanced half-cell reaction first.

Cr₂O₇^2- (aq) ------------> Cr³⁺ (aq)

Step-2 : Balance all atoms except O and H. This can be done by multiplying Cr³⁺ (aq) by 2.

Cr₂O₇^2- (aq) ------------> 2Cr³⁺ (aq)

Now Cr atom is balnced.

Step -3: Now we need to balance O and H. Since we need to balance O and H in basic medium, we need to achieve this by adding H2O and OH^-.

For each excess O - atom, add 1 H2O molecule on the same side and 2 OH^- ions on the opposite side.

O + H2O ---------> 2OH^-

For each excess H- atom, add 1 OH^- ions on the same side and 1 H2O molecule on the opposite side.

H + OH^-   ---------> H2O

Since there are 7 excess O- atoms on the left side, we need to add 7H2O on left and 14OH- ions on right side.

Cr₂O₇^2- (aq) + 7 H2O ------------> 2Cr³⁺ (aq) + 14 OH^-

Step-4: Now we need to balance electron. On the left side, there are 2 unit negative charge and on the right side 8 unit negative charge. Hence we need to add 6 e- on left. Hence the final balanced half-cell reaction is

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

Hence coefficient on water of the final balanced half-cell reaction is 7. Hence C is the correct answer.