Answer to Question #236586 in General Chemistry for Mickayla

Question

Calculate the cell potential for the following reaction:

Ni(s) + 2 Cu⁺(0.010M) → 2 Cu(s) + Ni⁺²(0.0010M)

1. 0.40 V
2. -0.43 V
3. 0.43 V
4. 0.34 V
5. 0.77 V

Answer 1

At anode (Oxidation reaction occur)

Ni (s)-------->Ni²⁺(aq) + 2e^-

E⁰cell= -0.25 V

At cathode Reduction reaction occur

2Cu⁺(aq) + 2e^- ------->2Cu(s).

E⁰cell= 0.520 V

So overall standard cell potential is

E^o_cell= E^ocell _reduction - E^ocell _oxidation

E^o_cell = 0.520-(-0.25)

=0.77V

E_cell =E^ocell - 0.059/n{log[Ni²⁺]/[Cu⁺]²}

n= no.of change of electron

E_cell=0.77-0.059/2{log[0.0010]/[0.010²]}

E_cell = 0.77-(0.0591/2)log(10)

E_cell=0.77-(0.0591/2)(1)

E_cell=0.77-0.0295

E_cell=0.7405V