Adding half cell potentials or reactions

You need to add it so it cancels out the electrons, right?
So take the following example...
Cu+ + e- = Cu E = 0.522V
Cu++ + 2e- = Cu E = 0.340V
So I would need to double the first reaction
2Cu+ + 2e- = 2Cu
But my question is, do the potential also doubles?
Like E = 1.044V now

though you may double the equation,
we do not double the voltage

the text book standard voltages given are for 1 Molar solutions...
to get double the electrons use twice as many ml of solution of the same 1 Molar

think of all of the different size 12 Volt car batteries
the voltages are all the same...
because the strength of the solution is the same,
the volume use varies
to get more energy

Cu+ + e- = Cu E = 0.522V
&
2Cu+ + 2e- = 2Cu E = 0.522V


========================

p.s. if you are settig up a voltaic cell
reduction: 2Cu+ + 2e- = 2Cu E = 0.522V
reduction: Cu++ + 2e- = Cu E = 0.340V

needs to be a redox:
reduction: 2Cu+ + 2e- = 2Cu E = 0.522V
oxidation: Cu ---> Cu++ + 2e- E = - 0.340V

your reaction:
2Cu+ & Cu ---> Cu++ & 2Cu E = 0.182
which simplifies to
2Cu+ ---> Cu++ & 1 Cu E = 0.182