Answer:
Reduction half-reaction: Cr₂O₇²- (aq) + 14H+ (aq) + 6e- ----> 2Cr³+ (aq) + 7 H₂O (l)
Oxidation half-reaction: 6 Cl- (aq) ---> 3 Cl₂ (g) + 6e-
Number of electrons = 6 electrons
Note: The question is incomplete. A similar but complete question is given below :
Write the balanced half-reactions for the following oxidation-reduction reaction that occurs in acid solution. Determine the number of electrons that appears in the balanced half reactions.
Cr₂0₇²- (aq) + CI- (aq) ---> Cr³+ (aq) + Cl₂(g)
Number of electrons =
Explanation:
A redox reaction is a reaction in which both oxidation and reduction occur simultaneously and to to the same extent.
Oxidation involves a loss of electron by atom of an element while reduction involve a gain of electrons.
A redox reaction can be split into two half-reaction; a reduction half-reaction andan oxidation half-reaction. The half-reactions in the given redox reaction is balanced as follows below:
Reduction half-reaction: Cr⁶+ being reduced to Cr³+
Cr₂O₇²- (aq) ---> Cr³+ (aq)
Since the reaction takes place in acidic medium, H+ and H₂O is added to the left and right-hand side of the equation respectively in order to balance the atoms
Cr₂O₇²- (aq) + 14H+ (aq) ----> 2Cr³+ (aq) + 7 H₂O (l)
Since 3 electrons are gained by Cr⁶+ in being reduced to Cr³+, 2 molecules of Cr⁶+ will accept 3 electrons each to give a total of 6 electrons
Cr₂O₇²- (aq) + 14H+ (aq) + 6e- ----> 2Cr³+ (aq) + 7 H₂O (l)
Oxidation half-reaction: Cl- ions are oxidized to Cl₂
Cl- (aq) ---> Cl₂
Two Cl- will each lose an electron to form Cl₂. Since the number of electrons gained in the reduction half-reaction is six, six electrons must be lost here as well. Therefore, the electrons involved and the number of moles of atoms involved are balanced as follows:
6 Cl- (aq) ---> 3 Cl₂ (g) + 6e-
