Asked by Katherine
The voltage generated by the zinc concentration cell described by,
Zn(s)|Zn2+ (aq, 0.100 M)||Zn2+ (aq, ____ M)|Zn(s)
is 24.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.
Hi, I posted this question awhile back but I think it got drowned out so I didn't want to comment on it again for fear that it'd go unseen.
DrBob told me the value should be above 0.1 M.. I tried again and I was wondering if this is correct:
0.024 = (8.314*298 K)/(2*96485)
0.024 = 0.012839 * ln[x/0.1]
x = 0.648 M?
thanks ahead of time!
Zn(s)|Zn2+ (aq, 0.100 M)||Zn2+ (aq, ____ M)|Zn(s)
is 24.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.
Hi, I posted this question awhile back but I think it got drowned out so I didn't want to comment on it again for fear that it'd go unseen.
DrBob told me the value should be above 0.1 M.. I tried again and I was wondering if this is correct:
0.024 = (8.314*298 K)/(2*96485)
0.024 = 0.012839 * ln[x/0.1]
x = 0.648 M?
thanks ahead of time!
Answers
Answered by
DrBob222
That looks ok to me although I seem to remember 0.7 something when I scratched this out a day or so ago but I've thrown my old worksheets away. I usually use
E = (-0.05916/2)log(0.1/x) but since you have changed the - sign in front to + then you have changed to x/0.1 and that makes it ok again.
E = (-0.05916/2)log(0.1/x) but since you have changed the - sign in front to + then you have changed to x/0.1 and that makes it ok again.
Answered by
Trevor
That answer is right
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