Here's the question:
Consider a voltaic electrochemical cell in which one compartment has a silver (Ag(s)) electrode with an electrolyte containing 1.00 M Ag+(aq). The other compartment has a chromium (Cr(s)) electrode with 1.00 M Cr3+(aq) ions. In this voltaic cell, _______ is produced at the anode, and _______ is produced at the cathode.
Ag+(aq) + e- <--> Ag(s) E0 = +0.80 V
Cr3+(aq) + 3e- <--> Cr(s) E0 = -0.74 V
Answers:
A) Ag(s), Cr3+
B) Ag+, Cr(s)
C) Ag+, Cr3+
D) Cr(s), Ag+
E) Cr3+, Ag(s)
The answer is E, Cr3+ for the first blank [produced at the anode], and Ag(s) for the second blank [produced at the cathode].
I knew that cathodes have the positive charges and anodes have the negative charges, which means that the electrons must be flowing from the cathode to the anode. That meant that Cr3+ or Ag+ had to be produced at the anode, which told me the answer was either B, C, or E. Then I knew that I could eliminate C because electrons were leaving the cathode, so it couldn't be Cr3+ because electrons are added in that reaction.
Left with answers B and E, I couldn't narrow down which one was which. I'm guessing that it has something to do with the E0 associated with each reaction (+0.80 V and -0.74 V, respectively).
Could anyone let me know if any of my logic in eliminating the first three parts was wrong, as well as giving me advice on how I can solve this problem (general advice is definitely 100% appreciated, but I'm specifically referring to how to choose between answers B and E).
Thanks very much for your help in advance.
Consider a voltaic electrochemical cell in which one compartment has a silver (Ag(s)) electrode with an electrolyte containing 1.00 M Ag+(aq). The other compartment has a chromium (Cr(s)) electrode with 1.00 M Cr3+(aq) ions. In this voltaic cell, _______ is produced at the anode, and _______ is produced at the cathode.
Ag+(aq) + e- <--> Ag(s) E0 = +0.80 V
Cr3+(aq) + 3e- <--> Cr(s) E0 = -0.74 V
Answers:
A) Ag(s), Cr3+
B) Ag+, Cr(s)
C) Ag+, Cr3+
D) Cr(s), Ag+
E) Cr3+, Ag(s)
The answer is E, Cr3+ for the first blank [produced at the anode], and Ag(s) for the second blank [produced at the cathode].
I knew that cathodes have the positive charges and anodes have the negative charges, which means that the electrons must be flowing from the cathode to the anode. That meant that Cr3+ or Ag+ had to be produced at the anode, which told me the answer was either B, C, or E. Then I knew that I could eliminate C because electrons were leaving the cathode, so it couldn't be Cr3+ because electrons are added in that reaction.
Left with answers B and E, I couldn't narrow down which one was which. I'm guessing that it has something to do with the E0 associated with each reaction (+0.80 V and -0.74 V, respectively).
Could anyone let me know if any of my logic in eliminating the first three parts was wrong, as well as giving me advice on how I can solve this problem (general advice is definitely 100% appreciated, but I'm specifically referring to how to choose between answers B and E).
Thanks very much for your help in advance.
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AN OX and a RED CAT!
OXdation happens at the ANode
REDuction happens at the CAThode
This is true regardless of the cell type (voltaic or electrolytic). Which ever half reaction has a greater E*, this is the one that will be reduced. The other reaction will happen in reverse (an oxidation) and so the E* becomes E*ox (opposite sign for the given E*)
Your E*cell is E*red+E*ox.
It MUST be positive because the question states the cell is voltaic, and all voltaic cells have a positive E*cell (and thus a negative G as G = -n f E)
Anyways, more positive E means reduction for that half reaction will occur, then it's AN OX RED CAT from there.
Hope that helps!
OXdation happens at the ANode
REDuction happens at the CAThode
This is true regardless of the cell type (voltaic or electrolytic). Which ever half reaction has a greater E*, this is the one that will be reduced. The other reaction will happen in reverse (an oxidation) and so the E* becomes E*ox (opposite sign for the given E*)
Your E*cell is E*red+E*ox.
It MUST be positive because the question states the cell is voltaic, and all voltaic cells have a positive E*cell (and thus a negative G as G = -n f E)
Anyways, more positive E means reduction for that half reaction will occur, then it's AN OX RED CAT from there.
Hope that helps!