Given the table below predict the numerical value of the standard cell potential for the reaction:
2 Cr(s) + 3 Cu2+(aq) 2 Cr3+(aq) + 3 Cu(s)
Half Reaction E (volts)
(1) Cr3+ + 3 e- Cr E= -0.74
(2) Cr3+ + e- Cr2+ E=-0.41
(3) Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
E= 1.33
(4) Cu+ + e- Cu E= 0.52
(5) Cu2+ + 2 e- Cu E= 0.34
(6) Cu2+ + e- Cu+ E= 0.16
how do i do this?
options:
a ) -0.4
b) 2.5
c) 1.08
d) 0.417
e) -1.08
(all units are volts)
i tried
E(products) - E(reactants) but it didn't work, i didn't get any of the options. :S
Well, well, well. It seems like you've got a bit of a pickle there. Don't worry, I'm here to help you out with a dash of humor. Let's spice up this electrochemical reaction, shall we?
To find the standard cell potential, you need to sum up the individual half-cell potentials, but you also need to remember to balance your equations.
Let's start with the oxidation half-reaction, where Cr(s) is being oxidized to Cr3+(aq). We need to double the given value of -0.74 V to account for the stoichiometry, giving us -1.48 V. Hang on tight, we're just getting started!
Now let's saunter on over to the reduction half-reaction, where Cu2+(aq) is being reduced to Cu(s). We'll use the given value of 0.34 V without any shenanigans.
Now, the trick here is to flip the oxidation potential and change its sign because we want to match the electrons in the half-reaction. So, -1.48 V becomes 1.48 V.
Finally, we add the reduction and oxidation potentials together (1.48 V + 0.34 V) to get a grand total of 1.82 V.
Voila! You've got your standard cell potential. Unfortunately, none of the given options match exactly. But hey, it's not uncommon for the right answer to evade us. Keep in mind, these options were made to clown around with your expectations. So, let's exit stage left, and give it another shot!
To determine the standard cell potential for the given reaction, you need to use the individual half-cell reactions and their corresponding standard cell potentials.
First, write down the given reaction and note the number of electrons transferred:
2 Cr(s) + 3 Cu2+(aq) 2 Cr3+(aq) + 3 Cu(s) (6 electrons are transferred)
Next, find the half-cell reactions that correspond to the species involved in the given reaction:
1) Cr3+ + 3 e- → Cr(s) E = -0.74 V (multiply by 2 to balance electrons)
2) Cu2+ + 2 e- → Cu(s) E = 0.34 V (multiply by 3 to balance electrons)
Now, sum up the individual half-cell reactions to obtain the overall balanced reaction:
2 Cr3+ + 6 e- → 2 Cr(s)
3 Cu2+ + 6 e- → 3 Cu(s)
Add the individual standard cell potentials of the half-cell reactions to calculate the overall standard cell potential:
(-0.74 V * 2) + (0.34 V * 3) = -1.48 V + 1.02 V = -0.46 V
Since the standard cell potential involves a reduction reaction, the value obtained is negative. Therefore, the numerical value of the standard cell potential for the given reaction is -0.46 V.
None of the options listed match exactly, but the closest one is option (d) 0.417 V.
To predict the numerical value of the standard cell potential for the given reaction, you need to use the principles of electrochemistry and the standard reduction potentials for each half-reaction. The standard cell potential, denoted as Ecell, can be calculated by subtracting the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction). Here's how you can determine the value:
1. Write the half-reactions for the given reaction:
Anode (Oxidation half-reaction): 2 Cr(s) - 6 e- -> 2 Cr3+(aq)
Cathode (Reduction half-reaction): 3 Cu2+(aq) + 6 e- -> 3 Cu(s)
2. Determine the reduction potentials for each half-reaction:
- The reduction potential for the anode half-reaction can be found in the given table. From the entries provided, the closest value is E= -0.41 V for the half-reaction Cr3+ + e- -> Cr2+.
- The reduction potential for the cathode half-reaction is given in the table as E= 0.34 V for the half-reaction Cu2+ + 2e- -> Cu.
3. Subtract the reduction potential of the anode from the reduction potential of the cathode:
Ecell = E(cathode) - E(anode)
= 0.34 V - (-0.41 V)
= 0.34 V + 0.41 V
= 0.75 V
The predicted numerical value of the standard cell potential for the given reaction is 0.75 volts. None of the provided options (a) -0.4, b) 2.5, c) 1.08, d) 0.417, e) -1.08) match this value exactly. If none of the options match, it is possible that there was an error in calculation or the options provided are not accurate.
First, you have no arrows. How can you possibly know which are products and which are reactants. I think you do yourself a disservice by not including an arrow.
For 2Cr + 3Cu^2+ ==> 3Cu + 2Cr^3+
Divide this into its two half cells.
Cr(s) ==> Cr^3+ + 3e
Cu^2+ + 2e ==> Cu(s)
Now look at the half cell voltages given to you. 1 is the one you want for Cr. That is
Cr3+ + 3 e- ==> Cr E= -0.74 which is just the reverse of your half cell. Therefore, the voltage for the reverse reaction is +0.74.
For the Cu, you want #5
Cu2+ + 2 e- ==> Cu E= 0.34
Eocell is Eox + Ered = 0.74+0.34 = ?