determining unknown quantity in cell

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

determining unknown quantity in cell

Determine the unknown quantity in the cell. E=0.050 V
Pb(s)|Pb^2+(aq,?)|Ni^2+(aq, 0.20 mol/L)|Ni(s)
Pb^2+(aq)+2e-→Pb(s)
E°=-0.13 V
Ni^2+(aq)+2e-→Ni(s)
E°=-0.23 V

Pb(s)→Pb^2+(aq)+2e-
Ni^2+(aq)+2e-→Ni(s)
e-s are cancelled out
Pb(s)+Ni^2+(aq)→Pb^2+(aq)+Ni(s)

E°cell=E°cath - E°anode
=-0.23V-(+0.13V)
=-0.10V

E=E°-(RT/nF)lnQ
0.050=-0.10-(8.31x10^-2)(298)/(2)(96485)lnQ
The answer to Q is supposedly 8.14x10^-6, but I keep getting the wrong answer. What am I doing wrong?

Justin Sarquiz 2F
Posts: 106
Joined: Fri Aug 30, 2019 12:15 am

Re: determining unknown quantity in cell

The R constant should be 8.314 J/(mol*K) instead of 0.0831.

ALegala_2I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

Re: determining unknown quantity in cell

For the units to cancel out correctly, you have to use R = 8.314 J/(mol*K). Whenever you need to decide what R to use, always check the units of the other values given.

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