To calculate the standard electromotive force (emf) for a galvanic cell, we can use the standard reduction potentials for the half-reactions that take place at each electrode.
Given:
- \( E^\circ \) for \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \) (reduction of magnesium) = \(-2.37 , \text{V}\)
- \( E^\circ \) for \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \) (reduction of silver) = \(0.80 , \text{V}\)
Step 1: Standard Reduction Potentials
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For the magnesium electrode (anode):
- The magnesium is oxidized: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)
- The corresponding reduction potential for this half-reaction (oxidation) is reversed:
- \( E^\circ_{\text{Mg}} = +2.37 , \text{V} \) (since we're considering the oxidation reaction)
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For the silver electrode (cathode):
- The silver is reduced: \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \)
- \( E^\circ_{\text{Ag}} = 0.80 , \text{V} \)
Step 2: Calculate the Standard EMF (\( E^\circ_{\text{cell}} \))
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Substituting in values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}} - E^\circ_{\text{Mg}} \]
\[ E^\circ_{\text{cell}} = 0.80 , \text{V} - (-2.37 , \text{V}) = 0.80 , \text{V} + 2.37 , \text{V} = 3.17 , \text{V} \]
Step 3: Determine the Direction of the Reaction
The positive \( E^\circ_{\text{cell}} \) value indicates that the galvanic cell reaction is spontaneous in the forward direction (favoring product formation). The greater the positive value of the emf, the more favorable the reaction is for proceeding in the forward direction.
Conclusion
- The standard emf (\( E^\circ_{\text{cell}} \)) of the galvanic cell is 3.17 V.
- The reaction will favor the forward (product) direction since the \( E^\circ_{\text{cell}} \) is positive. This means that the reaction is spontaneous under standard conditions.