I'm adding this to my previous question..in our textbook its says Ecell=Ecathode-Eanode. That's my values were smaller. But you added them. Will that make a difference? Should I do it the way you did it? Because they have to be bigger numbers? Thank you.
I add just because I think that is easier but you will notice I'm adding an oxidation half reaction to a reduction half reaction to arrive at Ecell.
Your text is subtracting a reduction half reaction from a reduction half reaction. It makes no difference how you do it because the answers should be the same. The Ecathode-Eanode is simpler for some people because it doesn't require that you change one of the half reactions to an oxidation one. For example, for the
Zn|Zn^2+(1M)||Cu^2+(1M)|Cu.
The redn Eocell for Zn = -0.762 in my book and this is the anode.
The redn Eocell for Cu = +0.345 and this is the cathode.
Ecathode - Eanode = ECu-EZn = 0.345-(-0.762) = +1.107 v which is exactly what I obtained by adding.
I added Zn oxidn = +0.762
to Cu redn = 0.345
0.762+0.345 = 1.107 v.
Why way you do it depends upon which is more comfortable for you. (I grew up with Eanode-Ecathode but we worked with oxidation tables and not reduction tables. Also we had a different formula for the half cell potentials. I've had to change how I do it to accompany the times. Frankly I think the IUPAC made a mistake in choosing the system they did but I'm one small ripple in a very large pond.:-).