Given a cell based on the spontaneous reaction
2AgCl(s) + Zn(s) ® 2Ag(s) + 2Cl– + Zn2+
If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should
increase by 0.06 V
increase by 0.18 V.
decrease by 0.06 V.
decrease by 0.18 V.
increase by 0.35 V.
i keep getting the wrong answer- none that are listed anyways.
i have put into half-rxn, then used
G=-nFE and the dG=-RT InK, but cannot get the right answer pls help. its for an assignment due tonight- aust time
got my questions meixed up!
I have tried this one by getting the 3 half-rxn and adding them up. then using E=E*-RT/nF InQ. i get close to an answer but not exact. i get 0.088V
Look up E* for Zn (as written). I find 0.763.
Look up AgCl(s) + e ==> Ag(s) + Cl^- (as written). I find 0.222
Add to find E*cell. I find 0.985
Then
Ecell = E*cell -(0.06/n)log(products/reactants).
Ecell = E*-(0.06/n)log [(Zn^+2)(Ag)^2(Cl^-)^2/(Zn)(AgCl)]
Now substitute for
n = 2
(Zn^+2) = 1 M
(Ag) = standard state = 1
(Cl^-) = 1 M
(Zn) = standard state = 1
(AgCl) = standard state = 1
Plugging all that in gives you the log term of 1, log 1 = 0 and Ecell is just E* = 0.985 v
Now to the other scenario.
(Zn^+2) = 1M
(Ag)(s) = standard state = 1
(Cl^-) = 0.001 (remember to square it).
(Zn) = standard state = 1
(AgCl) = standard state = 1.
So that part becomes
Ecell = E*cell - (0.06/2)log(1 x 10^-6)
Ecell = 0.985 - (0.03)(-6)
So the change is + 0.18 and that is one of the choices.
You may have been using 0.0592 which would give you 0.1776 which still rounds to 0.18v.
(
thank you so much. I stuffed up at the start. i had 3 half-rxn. i didn't see that there was one for AgCl- it all makes thank. thankyou for your patience, It appriciated. I had to cram this subject in to get this assignment done. we are not supposed to start this topic til monday which is when the assignment is due- not fun so again thankyou!
To determine the effect of changing the chlorine ion concentration on the cell voltage, we need to consider the Nernst equation. The Nernst equation allows us to calculate the cell potential at non-standard conditions.
The Nernst equation for the given reaction is:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the cell potential at non-standard conditions,
- E° is the standard cell potential,
- R is the ideal gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin,
- n is the number of moles of electrons transferred in the balanced equation,
- F is Faraday's constant (96485 C/mol),
- ln is the natural logarithm,
- Q is the reaction quotient.
To determine the effect of decreasing the chlorine ion concentration on the cell voltage, we need to calculate the reaction quotient (Q) at the two different chlorine ion concentrations.
At the initial concentration of chlorine ions (1 M), the reaction quotient (Q) is equal to the ratio of the product concentrations (raised to their stoichiometric coefficients) to the reactant concentrations (raised to their stoichiometric coefficients):
Q_initial = [Ag+]² / ([Cl-]² * [Zn²⁺])
At the decreased concentration of chlorine ions (0.001 M), the reaction quotient (Q) becomes:
Q_decreased = [Ag+]² / ([Cl-]² * [Zn²⁺])
Since the zinc ion concentration is kept constant at 1 M, [Zn²⁺] does not change in both cases. Therefore, the effect of changing the chlorine ion concentration only affects the reaction quotient (Q).
Now, calculate the reaction quotient (Q) for both cases:
Q_initial = [Ag+]² / (1² * 1) = [Ag+]²
Q_decreased = [Ag+]² / (0.001² * 1) = 10^6 * [Ag+]²
Next, calculate the change in cell potential (∆E):
∆E = E_final - E_initial
As the concentrations of other species (Ag+, Zn2+) are constant, their contributions to the cell potential remain the same. Therefore, we can ignore them while calculating the change (∆E).
Now, substitute the values into the Nernst equation:
∆E = E_final - E_initial = E° - (RT / nF) * ln(Q_decreased / Q_initial)
Since we are interested in the change in cell voltage, we only need to consider the natural logarithm of the ratio of reaction quotients (ln(Q_decreased / Q_initial)).
ln(Q_decreased / Q_initial) = ln((10^6 * [Ag+]²) / [Ag+]²) = ln(10^6) = 6 * ln(10)
Finally, substitute this value into the ∆E equation:
∆E = E_final - E_initial = E° - (0.0592 V / n) * (6 * ln(10))
Now, we can compare the final and initial cell potentials. If the cell voltage increases, then ∆E > 0. If the cell voltage decreases, then ∆E < 0.
Given the answer options, we can conclude that the cell voltage should decrease by 0.06 V.