To calculate the standard electromotive force (emf, \( E \)) of the galvanic cell consisting of a magnesium (Mg) electrode and a silver (Ag) electrode, we can use the following relationship:
\[ E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Step 1: Identify the electrodes
- Anode: Magnesium electrode (Mg)
- Cathode: Silver electrode (Ag)
Step 2: Determine the standard reduction potentials
From the data provided:
- For magnesium: \( E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 , \text{V} \) (Note: the value is negative because magnesium tends to oxidize, losing electrons.)
- For silver: \( E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 , \text{V} \)
Step 3: Write the half-reactions
-
Oxidation at the anode (Mg): \[ \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \]
-
Reduction at the cathode (Ag): \[ \text{Ag}^{+} + e^- \rightarrow \text{Ag} \]
Step 4: Calculate the standard emf of the cell
Using the standard reduction potentials in the formula, we can calculate \( E^\circ \):
\[ E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ = E^\circ_{\text{Ag}^{+}/\text{Ag}} - E^\circ_{\text{Mg}^{2+}/\text{Mg}} \] \[ E^\circ = 0.80 , \text{V} - (-2.37 , \text{V}) \] \[ E^\circ = 0.80 , \text{V} + 2.37 , \text{V} \] \[ E^\circ = 3.17 , \text{V} \]
Step 5: Determine the favored direction of the reaction
Since \( E^\circ = 3.17 , \text{V} \) is positive, the reaction favors the forward direction (toward the formation of products). In electrochemistry, a positive cell potential indicates that the reaction will proceed spontaneously as written, thus producing Mg²⁺ ions and solid silver.
Summary
- The standard emf of the galvanic cell is \( E^\circ = 3.17 , \text{V} \).
- The positive value of \( E^\circ \) indicates that the reaction favors the forward (product) direction.
Hence, the cell will function spontaneously, producing electrical energy from the electrochemical reaction between magnesium and silver ions in their respective solutions.