Activity 3: Based on the above example, determine the voltage of the voltaic cell combining the silver (Ag/Ag+) and copper (Cu/Cu2+) electrodes. a. First, look up the reduction potential of these two half reactions: Ag+(aq) + e---> Ag (s) E°red = ____________V Cu2+ (aq) + 2 e---> Cu (s) E°red = ____________V b. Now, which reduction is more prone to occurring? Ag/Ag+ or Cu/Cu2+? Circle the equation above. c. The other reaction, being less prone to happening, will thus go in the reverse direction. Write the equation of the reverse reaction below: → d. Now, to write the overall redox equation, one needs to balance out the electrons. This is done by multiplying the half reactions by integers such that the number of electrons being transferred is the same. Then add the two half-reactions together (the number of electrons on each side should cancel each other out). -- > e. What is the cell potential of the resulting electrochemical cell? Recall that you simply subtract from the reduction potential of the electrode undergoing reduction (what you circled for part b) with the potential of the electrode undergoing oxidation (part c). Reduction potential is an intensive property independent of amount, so it doesn’t matter how you combine the two halfreactions in part d, the voltages are still what you see on the table. Overall voltage is: _________V - _________________V = _____________V
Copy down the potential from the diagram above, as shown on the voltmeter: ___1.10_______ V
Note: Voltage (V) is a measurement of the
how charged particles are being pushed in one direction or another. In the case of the copper/zinc
electrochemical battery, the 1.10 V means that each Coulomb (C) of electric charge will gain an energy
of 1.10 Joule (J) (1 V = 1 J/C). For reference, 1 electron has a charge of 1.602 x 10-19 C (or 1 mole of
electrons carry about 100,000 C of electric carge).
The common data of electrochemical potential is in the form of standard reduction potential tables, like
the one shown below (similar to Table 17.1 in OpenStax):
****Note that the reduction of H+ to become H2 is set as 0.00 V. The electrode made of this combination is
called the standard hydrogen electrode (SHE), and standard reduction data is usually presented with
respect to SHE. One intuitive way to understand the standard reduction table is as follows: the higher
the reduction potential, the more prone the reduction half reaction is to occur. Thus, the reduction of
Cu2+ (aq) to Cu(s) with a potential of +0.34 V, is more likely to happen than the reduction of H+ to H2
(0.00 V). On the other hand, the reduction of Zn2+ (aq) to Zn(s) with a potential of -0.76 V is less prone to
occur than the SHE reduction. Thus, when we combine the copper (Cu/Cu2+) electrode with the zinc
(Zn/Zn2+) electrode, the copper reduction is more prone to happen by a voltage of 0.34 V – (-0.76 V) = 1.10 V. At the copper electrode, the reduction will happen. At the zinc electrode (less prone to
reduction), the oxidation will occur instead
Activity 3: Based on the above example, determine the voltage of the voltaic cell combining the silver
(Ag/Ag+) and copper (Cu/Cu2+) electrodes.
a. First, look up the reduction potential of these two half reactions:
Ag+(aq) + e---> Ag (s) E°red = ____________V
Cu2+ (aq) + 2 e---> Cu (s) E°red = ____________V
b. Now, which reduction is more prone to occurring? Ag/Ag+ or Cu/Cu2+? Circle the equation
above.
c. The other reaction, being less prone to happening, will thus go in the reverse direction. Write
the equation of the reverse reaction below:
→
d. Now, to write the overall redox equation, one needs to balance out the electrons. This is done
by multiplying the half reactions by integers such that the number of electrons being transferred
is the same. Then add the two half-reactions together (the number of electrons on each side
should cancel each other out).
-- >
e. What is the cell potential of the resulting electrochemical cell? Recall that you simply subtract
from the reduction potential of the electrode undergoing reduction (what you circled for part b)
with the potential of the electrode undergoing oxidation (part c). Reduction potential is an
intensive property independent of amount, so it doesn’t matter how you combine the two halfreactions in part d, the voltages are still what you see on the table.
Overall voltage is: _________V - _________________V = _____________V
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