Consider a voltaic cell constructed with a silver electrode placed in a 1.0 M AgNO3 solution and an aluminum electrode placed in a 1.0 M Al(NO3)3 solution. Write out the half reactions that occur at the anode and the cathode (4 points). Determine the cell potential and the direction of electron flow

2.46V

Explanation:

Oxidation half equation

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

Reduction half equation

3Ag^+(aq) +3e > 3Ag(s)

Since the concentration of each solution is 1M we assume standard conditions

E°cell = E°cathode - E°anode

E°cathode= 0.80V

E°anode= -1.66V

E°cell = 0.80 - (-1.66)

E°cell = 2.46V

b) nuclear ive learned that in a pedtrodic table