What is the correctly balanced reaction for a Li-Cr voltaic cell?
Andy, Jewel, Anna, etc etc. PLEASE use the same screen name. I can answer better and more fully if you use the same screen name throughout. It's confusing if I think I'm dealing with more than one.
You don't say where the Cr is coming from. I have assumed Cr metal and Cr^3+ and not [CrO4]^2- or [Cr2O7]^2-
Li(s) ==> Li^+ + e
Cr^3+ + 3e ==> Cr(s)
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Multiply equation 1 by 3 and add to equation 2 for the cell reaction.
To determine the balanced reaction for a Li-Cr voltaic cell, we need to consider the oxidation and reduction half-reactions.
1. Identify the oxidation half-reaction:
Li → Li+ + e-
2. Identify the reduction half-reaction:
Cr3+ + 3e- → Cr
3. Balance the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3:
3Li → 3Li+ + 3e-
4. Combine the two half-reactions and simplify if necessary:
3Li + Cr3+ → 3Li+ + Cr
Thus, the correctly balanced reaction for a Li-Cr voltaic cell is:
3Li + Cr3+ → 3Li+ + Cr
To determine the correctly balanced reaction for a Li-Cr voltaic cell, we need to consider the half-reactions occurring at the anode and cathode.
First, let's identify the standard reduction potentials (E°) for both lithium (Li) and chromium (Cr) half-reactions. These values can be found in tables of standard reduction potentials (commonly referred to as reduction potentials tables).
We have the following half-reactions:
1. Oxidation half-reaction at the anode: Li → Li+ + e- (reduction potential: E°1)
2. Reduction half-reaction at the cathode: Cr3+ + 3e- → Cr (reduction potential: E°2)
To construct a voltaic cell, the half-reactions need to be aligned in a way that allows the electrons released in the oxidation half-reaction to flow through an external circuit to reach the reduction half-reaction.
In order for this to occur, the half-reaction with the higher reduction potential will be reversed, ensuring that the electrons flow from the anode to the cathode.
Let's assume that E°1 > E°2, indicating that the reduction potential for the Li half-reaction is higher. We need to reverse the Li half-reaction:
3. Reversed oxidation half-reaction at the anode: Li+ + e- → Li
Now, we can balance the two half-reactions and combine them to form the overall balanced reaction for the Li-Cr voltaic cell:
Li + Cr3+ → Li+ + Cr
By canceling out the common species (Li+), the final balanced reaction becomes:
Li + Cr3+ → Cr
This balanced reaction represents the correctly balanced reaction for a Li-Cr voltaic cell. Remember, the actual voltage and reaction conditions may vary depending on the specific electrolyte used and other factors.