Question

can someone pleassseee help me with this chem question

Answer 1

so
Balance the following redox equation, identifying the element oxidized and the element reduced. Show all of the work used to solve the problem.

Ag + CN- + O2 yields Ag(CN)2- + H2O

Answer 2

Ag is oxidized and O is reduced

Answer 3

And i need help with the rest

Answer 4

Step 1:
Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product.

Answer 5

Step 2:
Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction.

Answer 6

Step 3:
Balance oxygen by adding H2O to one side of each half-reaction.

Answer 7

Step 4:
Balance hydrogen atoms. This is done differently for acidic versus basic solutions.

Answer 8

Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.

Answer 9

Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H3O+, H2O, or OH- appears on both sides of the final equation, cancel out the duplication also.

Answer 10

Finally balancing both sides for excess of H2O

Answer 11

On the left side, Ag has an oxidation number of 0 (it is an uncombined element). On the right side, Ag is +1. Why?

Ag = +1
+ 2CN- = -2
-----------------------------
Ag(CN) ion charge = -1

Note that oxygen on the left side also has an oxidation number of 0 for the same reason as Ag. But on the right side of the equation, it has formed H2O and has a charge of -2.

CN- is really a spectator ion (isn't oxidized or reduced) so we can ignore it for now.

Let's balance each half-reaction. I'm using = as an arrow sign.

Oxidation: Ag = Ag+ + e- (done)

Reduction: O2 = H2O
Put a 2 in front of H2O O2 = 2H2O
to balance oxygen
Put 4H+ on the left side O2 + 4H+ = 2H2O
to balance hydrogen

Put 4e- on the left to O2 + 4H+ + 4e- = 2H2O

To add the oxidation and reduction reactions together, I need to multiply the oxidation reaction by 4 so its electrons will cancel with the four that are in the reduction reaction.
Doing that you get

4Ag + O2 + 4H+ = 4Ag+ + 2H2O

In alkaline solution, you find the H+ above and add an equal number of OH- to BOTH sides of the equation: The 4H+ + 4OH- give 4 H2O.

4Ag + O2 + 4H+ = 4Ag+ + 2H2O
+4OH- + 4OH-
--------------------------------------…
4H2O

If I delete two H2O from each side I get:

4Ag + O2 + 2H2O = 4Ag+ + 4OH-

Finally, I can add 2CN- to each side for this:

4Ag + O2 + 2H2O + 8CN- = 4Ag(CN)- + 4OH-

Answer 12

Thank you! I will try it