Consider a galvanic cell based on the following line notation at 303 K: *The ans
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Consider a galvanic cell based on the following line notation at 303 K:
*The answer i have in the picture was marked incorrect, thanks!
Consider a galvanic cell based on the following line notation at 303 K: where the standard cell potential is 0.38 V. What will the following changes do to the potential of the cell? no change in potential of the cell decrease potential solution) no change in potential Add equal amounts of water to both the anode and cathode portions Remove water from the anode portion of the cell (concentrating the Decrease the mass of the Cu Add hint text here. Submit Answer Incorrect. Tries 2/3 Previous TriesExplanation / Answer
The Galvin cell is given as
Fe?Fe3+??Cu2+?Cu
The left hand side represents the anode (oxidation) while the right hand side represents cathode (reduction). The half cell reactions are
Fe (s) ---------> Fe3+ (aq) + 3 e-
Cu2+ (aq) + 2 e- ----------> Cu (s)
The balanced equation is
3 Cu2+ (aq) + 2 Fe (s) --------> 3 Cu (s) + 2 Fe3+ (aq)
The Nernst equation is given as
E = E0 + RT/6F*ln [Ox]/[Red]
where 6 denotes the number of electrons in the process. Here,
ln [Ox]/[Red] = ln [Fe3+]2/[Cu2+]3
where [..] denote molar concentrations. Now, we define molar concentration of a substance as
Molar concentration, [..] = (moles of substance)/(volume of solution).
When both the anode and the cathode compartments are diluted by adding equal amounts of water, then the volume of the solution increases or decreases by an equal amount and thus, the natural logarithm of the ratio of the molar concentrations, i.e,
ln [Fe3+]2/[Cu2+]3 will remain unchanged. Thus, E will remain unchanged, i.e, there is no change in potential of the cell.
When water is removed from the anode compartment, [Fe3+] increases (this is because the volume of the solution in the anode compartment decreases). Therefore,
ln [Fe3+]2/[Cu2+]3
will increase and thus, E will increase. Therefore, the removal of water from the cell compartment increases the cell potential.
When the mass of Cu is decreased, the moles of Cu = (mass of Cu)/(molar mass of Cu) decreases and hence, the molar concentration of Cu decreases. Therefore,
ln [Fe3+]2/[Cu2+]3
increases and thus, E increases. Therefore, removing Cu from the cathode compartment increases the cell potential.
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