If you had written them as half cells you would have figured out what to do.
2F^- ==> F2 + 2e
AuCl4^- + 3e ==> Au + 4Cl^-
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Now multiply the fluoride equation by 3 and the Au equation by 2 (to obtain 6e lost and 6e gained) and it will balance.
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?
4 answers
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.