Question

A galvanic cell consists of a Mg electrode in a 1 M Mg(NO3)2 solution and another metal electrode X in a 1 M X(NO3)2 solution.

The galvanic cell has an E°cell value of 1.61 V. Which of the following elements fits the identity of X. (Use table table 18.1)



Select one:

a.
Pb


b.
Zn


c.
Ni


d.
Fe


e.
Mn

Answer 1

To determine the identity of metal X, we need to compare the standard reduction potentials of the possible metals with the standard reduction potential of the Mg half-reaction.

From Table 18.1, we can find the standard reduction potentials for each of the metals listed:

Pb: -0.13 V

Zn: -0.76 V

Ni: -0.25 V

Fe: -0.44 V

Mn: -1.18 V

The reduction half-reaction for the Mg electrode is:

Mg2+ + 2e- → Mg E° = -2.37 V

The overall reaction for the galvanic cell is:

Mg(s) + X2+(aq) → Mg2+(aq) + X(s)

The standard cell potential is given by:

E°cell = E°(cathode) - E°(anode)

where the cathode is the reduction half-reaction and the anode is the oxidation half-reaction.

Substituting the given values, we get:

1.61 V = E°(X2+/X) - (-2.37 V)

Simplifying, we get:

E°(X2+/X) = 1.61 V + 2.37 V = 3.98 V

Comparing E°(X2+/X) with the standard reduction potentials in Table 18.1, we see that only zinc (Zn) has a reduction potential that is more negative than 3.98 V. Therefore, the metal X is zinc (Zn).

Therefore, the answer is (b) Zn.