For the reaction: Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) E°cell = 0.31 V Imagine a Voltaic cell is set up to investigate the effects of different changes of conditions on cell potential. Predict the outcome of the following changes with Ecell > E°cell or Ecell < E°cell or no change. a) Increase the mass of the Lead cathode b) [Fe2+] 0.10 mol L-1; [Pb2+] 0.0001 mol L-1 c) [Pb2+] 1.5 molL-1, [Fe2+] 0.00001 mol L-1 d) Reduce the mass of the Iron anode

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Chapter1: Chemical Foundations
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For the reaction: Pb2+(aq) + Fe(s) → Pb(s) +
Fe2+(aq) E°cell = 0.31 V Imagine a Voltaic cell
is set up to investigate the effects of different
changes of conditions on cell potential.
Predict the outcome of the following changes
with Ecell > E°cell or Ecell < E°cell or no
change.
a) Increase the mass of the Lead cathode
b) [Fe2+] 0.10 mol L-1; [Pb2+] 0.0o01 mol L-1
c) [Pb2+] 1.5 mol L-1; [Fe2+] 0.00001 mol L-1
d) Reduce the mass of the Iron anode
Transcribed Image Text:For the reaction: Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) E°cell = 0.31 V Imagine a Voltaic cell is set up to investigate the effects of different changes of conditions on cell potential. Predict the outcome of the following changes with Ecell > E°cell or Ecell < E°cell or no change. a) Increase the mass of the Lead cathode b) [Fe2+] 0.10 mol L-1; [Pb2+] 0.0o01 mol L-1 c) [Pb2+] 1.5 mol L-1; [Fe2+] 0.00001 mol L-1 d) Reduce the mass of the Iron anode
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