Question

Represent the cell in which the following reaction takes place:
\text{Mg(s)} + 2Ag^{+} (0.001 M) \rightarrow \text{Mg}^{2+} (0.100 M) + 2Ag(s)
Calculate E_{cell} if E^\circ_{cell} = 3.17 V. (log 10 = 1)

Hint:

The Nernst equation can be used to calculate the cell potential based on concentration changes.

Answer 1

For the given reaction, we first write the two half-reactions: - Oxidation half-reaction: Mg(s) \rightarrow \text{Mg}^{2+} (0.100 M) + 2e^{-} - Reduction half-reaction: 2Ag^{+} (0.001 M) + 2e^{-} \rightarrow 2Ag(s) Now, to calculate the cell potential, we use the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] Here, \( n = 2 \) (the number of electrons transferred) and \[ Q = \frac{[\text{Mg}^{2+}][\text{Ag}]^2}{[\text{Ag}^{+}]^2[\text{Mg}]} \] Substituting the given values: \[ Q = \frac{(0.100)(1)^2}{(0.001)^2(1)} = 10^4 \] Now, apply the Nernst equation: \[ E_{\text{cell}} = 3.17 - \frac{0.0592}{2} \log 10^4 = 3.17 - \frac{0.0592}{2} \times 4 = 3.17 - 0.1184 = 3.0516 \, \text{V} \] Thus, the cell potential is \( E_{\text{cell}} = 3.0516 \, \text{V} \).