2co3+(aq)+2cl−(aq)→2co2+(aq)+cl2(g) . e∘=0.483 v what is the cell potential at 25 ∘c if the concentrations are [co3+]= 0.717 m , [co2+]= 0.363 m , and [cl−]= 0.152 m and the pressure of cl2 is pcl2= 3.40 atm ?

Look for Q. The equation you would use to find Q would be - [Co2+] ^2 [PCl2] / [Co3+] ^2 [Cl-]^2 
then plug in the the numbers you are given into the equation [0.363 M]^2 [3.40 atm] / [0.717 M] ^2 [0.152] ^2 0.4480 / 0.0119 = 37.7183 
Next you would want to determine the value of n = 2 
then you will want to plug in the values you have to the final equation E=E∘−(0.0591(n))logQ 
E =0.71 - (0.0591(2)) * log 37.7
E= 0.71 - 0.02955 * 1.58 
E= 0.71 – 0.0467
E= 0.6633

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer