Question

Which of the following would produced the highest voltage in a galvanic cell?

Select one:

a. 2Cr(s) + 3Ni²⁺(aq) → 2Cr³⁺ (aq) + 3Ni(s)

b. Li(s) + Ag⁺(aq) → Li⁺ (aq) + Ag(s)

c. Mg(s) + Zn²⁺(aq) → Mg²⁺ (aq) + Zn(s)

d. Na(s) + K⁺(aq) → Na⁺ (aq) + K(s)

Answer 1

B) Li(s)+Ag+(aq)=> Li+(aq)+ Ag (s)

The half cell reaction with less reduction potential will act as anode and the other cell with more reduction potential will act as cathode. This is because ,the reduction potential shows how much the cell is reduced and reduction occurs at cathode . So the cell with more reduction potential will act as cathode .

First we will write the given equations in half cell reaction.

Then we will find the reduction potential value of all the half cell .

Then we divide the cells as anode and cathode .

Then we will calculate standard cell potential :

Ecell = E°cathode-E°anode

The cell with highest cell potential produces high voltage .

A)

Cr(s)=> Cr3+(aq) + 3e- ........E°red = -0.74V( anode )

Ni2+(aq) + 2e- => Ni(s) .....E°red= -0.23V (cathode)

Ecell= E°cathode - E°anode = -0.23V-(-0.74V) =0.51 V

B)

Li(s)=> Li+(aq) + e- .....E°= -3.04V (anode )

Ag+(aq)+ e-=> Ag (s)....E°= 0.80 V (cathode)

Ecell= 0.80V-(-3.04V) = 3.84 V

C)

Mg(s) => Mg2+(aq) + 2e- .......E°= -2.37V(anode)

Zn2+(aq) +2e-=> Zn (s) ......E°= -0.76V (cathode)

Ecell= -0.76V -(-2.37V) = 1.61 V

D)

Na (s)=> Na+(aq)+e-.....E°= -2.71 V (cathode)

K+(aq) +e-=> K(s) ......E°= -2.925 V(anode )

Ecell= -2.71-(-2.925)= 0.215 V

B has largest voltage in galvanic cell .