Using the reactivity series of metals, classify which would be the cathode in a reaction involving these two metals as electrodes.

Na or Ca?

Calcium will be the anode and sodium will be the cathode

Explanation:

I know that your confusion stems from the relative positions of the two metals in the activity series but I want you to get something clearly. Sodium metal is more reactive than calcium metal but calcium ion is more reactive (a better reducing agent) than the sodium ion.

Hence, in an electrochemical cell, a better reducing agent will function as the anode. This is why calcium is the anode in the cell mentioned in the question. Calcium ion has a reduction potential of -2.87 V while sodium ion has a reduction potential of -2.71 V, hence the answer.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally