00:01
Starting with the solution, here we have cu2 +, plus zn, which react to give cu +, zn2+.
00:15
We have given concentration of cu2 +, 0 .788 molar cu2 +, and 7 .32 is the concentration of zn2 +, 7 .32 molar zn2 +, concentrations are given.
00:42
We need to find out the cell voltage for this reaction.
00:50
So here the oxidation reaction would be zn gives zn2 +, plus 2 electron and reduction reaction would be cu2 +, accepts 2 electron and gives cu.
01:20
We need to use reduction potential table for the e0 values.
01:28
For oxidation it would be minus 0 .76 and here for reduction of cu2 +, the value is 0 .34.
01:41
Now we can calculate e cell by using equation e0 cell equals to e0 reduction minus e0 oxidation.
01:58
Now let's plug the values we have 0 .34 minus minus 0 .76.
02:07
On calculating we will obtain the value 1 .1 volt...