A galvanic cell is connected to a small light bulb for 40.0 minutes. The electrodes are
copper and silver and the copper electrode is immersed in a solution of copper sulfate.
During that time, the average current drawn by the bulb is 0.12 A.
a. What is more reactive, copper or silver?
b. Which is the anode and which is the cathode?
c. How many grams of copper dissolve from the (anode or cathode) during this
time?
d. If copper metal is ‘moving into solution’, is this an oxidation or reduction
a. I don't know what you mean with "more reactive" but the Eo oxidations values are shown below. Also I assume the Ag electrode is immersed in a silver salt of some kind. I presume you would say Cu is the more reactive metal.
Cu ==> Cu^2+ + 2e Eo = -0.344 v
Ag ==> Ag^+ + e Eo = -0.8
b. The anode is defined as the electrode at which oxidation takes place. That is Cu.
d. See b for the answer.
c. Coulombs = amperes x seconds = ?= approx 300 but you need to do it more accurately. Then 96,485 coulombs will dissolve 63.54/2 grams Cu. So how much will dissolve with 300 coulombs. That will be
63.54/2 x (approx 300/96,485) = ?
thanks!
To answer these questions, we need to understand some basic concepts of galvanic cells and electrochemistry.
A galvanic cell consists of two electrodes, an anode (where oxidation occurs) and a cathode (where reduction occurs), that are connected by an electrolyte solution. In this case, the copper and silver electrodes are immersed in a solution of copper sulfate.
a. To determine which metal is more reactive, we need to refer to the activity series of metals. The activity series is a list of metals in order of their reactivity, with the most reactive at the top and the least reactive at the bottom.
In the activity series, copper is above silver, which means copper is more reactive than silver. So, the answer is copper is more reactive than silver.
b. To determine which metal is the anode and which is the cathode, we first need to identify the half-reactions that occur at each electrode.
The half-reaction at the anode involves oxidation, where copper atoms lose electrons to form copper ions. This can be represented as:
Cu(s) -> Cu2+(aq) + 2e-
The half-reaction at the cathode involves reduction, where silver ions gain electrons to form silver atoms. This can be represented as:
Ag+(aq) + e- -> Ag(s)
Based on these half-reactions, we can conclude that copper is the anode (where oxidation occurs) and silver is the cathode (where reduction occurs).
c. To determine the amount of copper that dissolves from the anode during this time, we need to use Faraday's law of electrolysis. According to Faraday's law, the amount of substance consumed during electrolysis is directly proportional to the amount of electrical charge passed through the cell. The formula is:
m = (Q * M) / (n * F)
Where:
m = mass of substance consumed (in grams)
Q = total charge passed through the cell (in Coulombs)
M = molar mass of the substance (in grams/mol)
n = number of electrons transferred in the balanced half-reaction
F = Faraday's constant (96500 C/mol)
To apply this formula, we need to know the charge passed (Q) and the number of electrons transferred (n). Since we are given the average current (0.12 A) and the time (40 minutes), we can calculate the charge passed using the formula:
Q = I * t
Where:
I = current (in Amperes)
t = time (in seconds)
First, convert 40 minutes to seconds:
40 minutes * 60 seconds/minute = 2400 seconds
Then, calculate the charge passed:
Q = 0.12 A * 2400 seconds = 288 Coulombs
Now, we need to determine the number of electrons transferred in the balanced half-reaction for copper. By looking at the balanced half-reaction (Cu(s) -> Cu2+(aq) + 2e-), we can see that 2 electrons are transferred for every copper atom consumed.
Plugging these values into the formula, we get:
m = (288 C * 63.5 g/mol) / (2 mol * 96500 C/mol)
Simplifying:
m = 0.009 g
Therefore, approximately 0.009 grams of copper dissolve from the anode during this time.
d. If copper metal is "moving into solution," it means that copper atoms are losing electrons to form copper ions (Cu -> Cu2+). This process corresponds to the anode, where oxidation occurs. Oxidation is the loss of electrons, so if copper is being oxidized, this process is an oxidation reaction.
In summary:
a. Copper is more reactive than silver.
b. Copper is the anode, and silver is the cathode.
c. Approximately 0.009 grams of copper dissolve from the anode.
d. The process of copper "moving into solution" is oxidation.