Question

In: Chemistry

Consider a cell based on the following line notation at 298 K: Cu | Cu2+ (0.237...

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

Solutions

Expert Solution

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


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