Calculate the voltage produced by the cell Sn(s)|Sn2+|Ag(s) at 25°C given: [Sn2+]=0.15 M [Ag+]=1.7 M

The galvanic cell consists of a hydrogen half-cell and a magnesium half-cell. The standard reduction potential for the hydrogen half-cell (Standard Hydrogen Electrode, SHE) is defined as $0.00 \, V$.

$$2H^+(aq) + 2e^- \to H_2(g) \quad E^\circ = 0.00 \, V$$

Magnesium is a more reactive metal than hydrogen, indicating it has a greater tendency to be oxidized. Therefore, the magnesium electrode will act as the anode (where oxidation occurs), and the hydrogen electrode will act as the cathode (where reduction occurs).

Anode (oxidation):

$$Mg(s) \to Mg^{2+}(aq) + 2e^-$$

Cathode (reduction):

$$2H^+(aq) + 2e^- \to H_2(g)$$

Adding the anode and cathode half-reactions:

$$Mg(s) + 2H^+(aq) \to Mg^{2+}(aq) + H_2(g)$$

The cell potential ($E_{cell}$) is given as 2.36 V. The formula for the standard cell potential is:

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$

Substituting the known values:

$$2.36 \, V = 0.00 \, V - E^\circ_{Mg^{2+}/Mg}$$

Solving for $E^\circ_{Mg^{2+}/Mg}$:

$$E^\circ_{Mg^{2+}/Mg} = \boxed{\mathbf{-2.36 \, V}}$$