See Figure 3, describe how the values of the electromotive force with respect to
ID: 522219 • Letter: S
Question
See Figure 3, describe how the values of the electromotive force with respect to the standard hydrogen electrode are determined. How can one use this table to compare metals and determine whether or not one metal will corrode with respect to the other?
Two outcomes Corrosion Electrodeposition H2(gas) 2e 2e ne ne Mn+ Mn+ ions IOns as 25 C 25' C 1 M Mn+ so n 1 M H+ sol' n 1 M n sol n 1 M H sol' n Metal is the anode Metal is the cathode meta 0 relative to Pt) metal 0 (relative to Pt) Adapted from Fig. 16.2, Standard Electrode Potentia Calister & Rethwisch 4e. Figure 3.Explanation / Answer
The potential difference between the two electrodes in a galvanic cell is called an electromotive force or cell potential of the cell. It is measured in volts.
Cell potential = Potential of the half cell at Cathode - Potential of the half cell at Anode
E0cell = E0cathode - E0anode
In the first cell, metal is the anode while the platinum electrode is the cathode. Note, reduction takes place at cathode while oxidation takes place at anode. At anode, the metal M(0) is oxidised to Mn+ by losing electrons. These electrons flow to the cathode and reduce H+ ions to H2. The half reactions can be given as:
2H+(aq) + 2e- ------> H2(g)…………………….Reduction
M(s) -----> M+n(aq) + ne-…………..Oxidation
2nH+(aq) + 2ne- + 2M(s) -----> nH2(g) + 2M+n(aq) + 2ne-
Overall reaction is given as
2nH+(aq) + 2M(s) -----> nH2(g) + 2M+n(aq)
From the standard reduction potential table, we know that E0red( H2) = 0 V
E0cell = E0cathode - E0anode
E0cell = - E0anode
The net cell potential will be negative and the reaction is non-spontaneous. At anode we will see metal dissolving or depleting while at cathode we will see bubbles of H2 gas. In other words the metal gets corroded.
In the second cell, metal is the cathode while the platinum electrode is the anode. At anode, the H2 gas is oxidised to H+ by losing electrons. These electrons flow to the cathode and reduce Mn+ ions to M(0). The half reactions can be given as:
H2 (g) -----> 2H+ + 2e-…………..Oxidation
M+n(aq) + ne- -----> M(s) …………..Reduction
2M+n(aq) + 2ne- + nH2 (l) -----> 2M(s) +2nH+ + 2ne-
Overall reaction is given as
2M+n(aq) + nH2 (l) -----> 2M(s) +2nH+
From the standard reduction potential table, we know that E0red( H2) = 0 V
E0cell = E0cathode - E0anode
E0cell = E0cathode
The net cell potential will be positive and the reaction is spontaneous. At cathode we will see more metal depositing. In other words, the metal is electroplated.
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