To calculate the standard electromotive force (E°) of the galvanic cell consisting of a magnesium (Mg) electrode and a silver (Ag) electrode, we will use the standard reduction potentials of the half-reactions involved.
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Reduction half-reactions:
For magnesium: \[ \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \quad (E^\circ = -2.37 , \text{V}) \]
For silver: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^\circ = +0.80 , \text{V}) \]
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Identify which half-reaction is undergoing oxidation and which is undergoing reduction:
- The Mg half-reaction is on the left side (being oxidized), so it will have a negative electrode potential (serving as anode).
- The Ag half-reaction is on the right side (being reduced), with a positive electrode potential (serving as a cathode).
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Standard emf (E°) of the cell is calculated using: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Plugging in our values for the respective half-reactions: \[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}} - E^\circ_{\text{Mg}} = 0.80,V - (-2.37,V) \] \[ E^\circ_{\text{cell}} = 0.80,V + 2.37,V = 3.17,V \]
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Determine the direction the reaction favors:
- A positive E° value indicates that the spontaneous reaction favors the forward direction (towards product formation).
- Given that \( E^\circ_{\text{cell}} = 3.17,V \) (which is positive), the cell reaction will favor the forward direction.
Conclusion:
The standard emf (E°) of the galvanic cell is \( 3.17 , \text{V} \), indicating that the reaction will favor the forward (product) direction.