For a voltaic cell consisting of Al(s) in Al(NO3)3(aq) and Cu(s) in Cu(NO3)2(aq), what is Ecell, given [Al3 ]

2.0 V

Explanation:

For the oxidation half cell;

Al(s) > Al^3+(aq) + 3e.

For reduction half cell;

Cu^2+(aq) +2e > Cu(s).

E°cell = E°cathode - E°anode

But;

E°cathode= 0.34 V

E°anode = -1.66 V

E°cell= 0.34 -(-1.66)

E°cell= 2.0 V

chemistry sucks a**

explanation:

the temperature of the gas, in k, not celsius.