I think these I know how to do ....just want someone to check..still sorta mixed up on the first question. Thanx
For the following reaction, what is the correctly balanced reduction half-reaction?
Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + OH-(aq)
A. e- + Fe(OH)3(s) → Fe(s) + OH-(aq)
B. 3 e- + Fe(OH)3(s) → Fe(s) + 3 OH-(aq)
C. Fe(OH)3(s) → Fe(s) + 3 OH-(aq) + 3 e-
D. Cu(s) → Cu+(aq) + e-
I think answer is C
For the following reaction, what is the correctly balanced overall reaction?
Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + OH-(aq)
A. Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + 3 OH-(aq)
B. Fe(OH)3(s) + 2 Cu(s) → Fe(s) + 2 Cu+(aq) + OH-(aq)
C. Fe(OH)3(s) + 3 Cu(s) → Fe(s) + 3 Cu+(aq) + 3 OH-(aq)
D. 3 Fe(OH)3(s) + Cu(s) → 3 Fe(s) + Cu+(aq) + 9 OH-(aq)
i thnk answer is A
if either is wrong can sum1 explain why? thanx
For the following reaction, what is the correctly balanced reduction half-reaction?
Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + OH-(aq)
To be reduced the electrons must be added which means on the left.
A. e- + Fe(OH)3(s) → Fe(s) + OH-(aq)
A can't be right because OH^- doesn't balance.
B. 3 e- + Fe(OH)3(s) → Fe(s) + 3 OH-(aq)
C. Fe(OH)3(s) → Fe(s) + 3 OH-(aq) + 3 e-
C can't be right because the electrons are on the right AND because the charge doesn't balance (zero on left and -6 on right).
D. Cu(s) → Cu+(aq) + e-
D can't be right because the electrons are on the right side
I think answer is C
For the following reaction, what is the correctly balanced overall reaction?
Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + OH-(aq)
A. Fe(OH)3(s) + Cu(s) → Fe(s) + Cu+(aq) + 3 OH-(aq)
A can't be right because the charge doesn't balance (zero on left and -2 on right--also the exchange of electrons is not the same).
B. Fe(OH)3(s) + 2 Cu(s) → Fe(s) + 2 Cu+(aq) + OH-(aq)
B can't be right because the charge doesn't balance and OH doesn't balance and electron transfer doesn't balance.
C. Fe(OH)3(s) + 3 Cu(s) → Fe(s) + 3 Cu+(aq) + 3 OH-(aq)
D. 3 Fe(OH)3(s) + Cu(s) → 3 Fe(s) + Cu+(aq) + 9 OH-(aq)
D can't be right because charge doesn't balance and because electron transfer doesn't balance.
i thnk answer is A
For the first question, you are correct. The correct balanced reduction half-reaction is option C: Fe(OH)3(s) → Fe(s) + 3 OH-(aq) + 3 e-.
To determine the balanced reduction half-reaction, you need to consider the change in oxidation states of the elements involved. In this case, the oxidation state of Fe changes from +3 in Fe(OH)3 to 0 in Fe, indicating a reduction. The number of OH- ions increases from 0 to 3, and 3 electrons are gained. Therefore, option C is correct.
For the second question, the correct answer is actually option B: Fe(OH)3(s) + 2 Cu(s) → Fe(s) + 2 Cu+(aq) + OH-(aq).
To determine the balanced overall reaction, you need to balance both the reduction half-reaction and the oxidation half-reaction. The reduction half-reaction is the one you already determined in the first question: Fe(OH)3(s) → Fe(s) + 3 OH-(aq) + 3 e-.
Now, let's consider the oxidation half-reaction. Copper (Cu) changes from 0 oxidation state to +1 in Cu+(aq). Therefore, the balanced oxidation half-reaction is: Cu(s) → Cu+(aq) + e-.
To balance the overall reaction, you need to ensure that the number of electrons gained in the reduction half-reaction is the same as the number of electrons lost in the oxidation half-reaction. In this case, that means multiplying the reduction half-reaction by 3, and the oxidation half-reaction by 2.
By doing that, you get:
3 Fe(OH)3(s) + 2 Cu(s) → 3 Fe(s) + 2 Cu+(aq) + 6 OH-(aq)
Therefore, the correct balanced overall reaction is option B: Fe(OH)3(s) + 2 Cu(s) → Fe(s) + 2 Cu+(aq) + OH-(aq).
I hope this explanation clarifies any confusion.