In: Chemistry
Consider a cell based on the following line notation at 298 K:
Cu | Cu2+ (0.237 M) || Ni2+ (1.29 M) | Ni
Given:
Cu2+ + 2 e- →
Cu Eo = 0.34 V
Ni2+ + 2 e- →
Ni Eo = -0.23 V
How many of the following responses are true?
1. Decreasing the concentration of the Cu2+ will
increase the concentration of the reaction
2. Using a Pt electrode in place of the Cu electrode will not
change the potential of the cell
3. Oxidation takes place at the anode
4. Adding equal amounts of water to both half reactions will
decrease the potential of the cell
5. Increasing the concentration of the Ni2+ will
increase the potential of the cell
Cu(s) + Ni2+ = Ni(s) + Cu2+(aq)
Q = [Cu2+] / [Ni2+]
1. Decreasing the concentration of the Cu2+ will increase the
concentration of the reaction. since cu2+ is being
oxidized, then decrease in Cu2+ will not favour the
equilibrium
2. Using a Pt electrode in place of the Cu electrode will not
change the potential of the cell FALSE, it will, since Pt
is another metal, has another value for reduction
potential
3. Oxidation takes place at the anode True, anode will
always carry on oxidation
4. Adding equal amounts of water to both half reactions will
decrease the potential of the cell false, since both are
2:2 ratio, so the concentrations cancel out according to Q =
[Cu2+]/[Ni].
5. Increasing the concentration of the Ni2+ will increase the
potential of the cell Increasing Ni2+, the species being
reduec, will favour the E potential, therefore, E
increases