Question
Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.
Pt(s) | F-(aq) | F2(g) || Cl-(aq), AuCl4-(aq) | Au(s)
I'm having trouble balancing this...
I get AuCl4- + 2F- --> Au + 4Cl- + F2
...but the charges don't cancel out on both sides. The left side has 3- and the right side has 4-. I've been trying to change the numbers around but then I upset the balanced atoms. Help?
Pt(s) | F-(aq) | F2(g) || Cl-(aq), AuCl4-(aq) | Au(s)
I'm having trouble balancing this...
I get AuCl4- + 2F- --> Au + 4Cl- + F2
...but the charges don't cancel out on both sides. The left side has 3- and the right side has 4-. I've been trying to change the numbers around but then I upset the balanced atoms. Help?
Answers
If you had written them as half cells you would have figured out what to do.
2F^- ==> F2 + 2e
AuCl4^- + 3e ==> Au + 4Cl^-
===========================
Now multiply the fluoride equation by 3 and the Au equation by 2 (to obtain 6e lost and 6e gained) and it will balance.
2F^- ==> F2 + 2e
AuCl4^- + 3e ==> Au + 4Cl^-
===========================
Now multiply the fluoride equation by 3 and the Au equation by 2 (to obtain 6e lost and 6e gained) and it will balance.
Thanks a lot! I forgot you had to write the half reactions first.
HI i WAS WONDERING, which
one was the anode?
one was the anode?
@Raffi
The Anode is always written first in this notation.
The Anode is always written first in this notation.
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