A concentration cell is constructed by placing identical Co electrodes in two Cu

Question

A concentration cell is constructed by placing identical Co electrodes in two Cu2 + solutions. Cu/Cu^2+ (0.002 M) || Cu^2 + (1 M)/Cu If the concentrations of the two Cu2 + solutions are 1.0 M and 0.002 M, calculate the potential of the cell. 0.080 V 0.030 V 0.020 V 1.2 V 1.0 V In voltaic cells, the salt bridge. is not necessary in order for the cell to work allows charge balance to be maintained in the cell acts as a mechanism to allow mechanical mixing of the solutions is tightly plugged with firm agar gel through which ions cannot pass drives free electrons from one half-cell to the other In the following voltaic cell: Zn/Zn^2 + (aq) || Cu^2 + (aq)/Zn the electrodes are connected by means of a membrane Zn/Zn^2 + is the cathode Zn/Zn^2 + is the anode the electrodes are connected by means of a membrane none of the above How many electrons flow in the following battery: Fe/Fe^2+ (aq) Au^3 + (ag)/Au 3e- 2e 5e- 6e zero Calculate Delta E for the following voltaic cell: Fe/Fe^2 + (10^-6 M) || Au^3 + (1M)/Au Knowing that Delta E^0 = + 2.00V -0.5 V 2.00 V 1.94 V 7V

Explanation / Answer

16)

Cu is both oxidised and reduced

Cu2+(cathode) + Cu -----> Cu2+(anode) + Cu

E = Eo - 0.059/2 log {[Cu2+]anode / [Cu2+]cathode}
[Cu2+]anode < [Cu2+]cathode to make E positive

anode will be the one with lower concentration

Since both cathode and anode is Zn,
Eo cell = Eo cathode - Eo anode = 0

E = Eo - 0.059/2 log {[Cu2+]anode / [Cu2+]cathode}
E = 0 - 0.059/2 log (0.002/1)
E = - (0.059/2 )*( - 2.699)
= 0.080 V
Answer: A

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