Question

What is the cell potential of the following voltaic cell if the
Ksp of Ag_{2}SO_{4} is 1.4x10−5?

Cu(s) | Cu+(1.0 M) || Ag+(Saturated
Ag_{2}SO_{4}) | Ag(s)

The answer is 0.19 V , please explain how and step by step.

Answer #1

First, with the Ksp value, let's calculate concentration of the
Ag^{+} and write an ICE chart:

r: Ag_{2}SO_{4(s)} --------> 2Ag^{+} +
SO_{4}^{2-}

i: s 0 0

e: 2s s

1.4x10^{-5} = (2s)^{2} * s

1.4x10^{-5} = 4s^{3}

s = (1.4x10^{-5}/4)^{1/3}

s = 0.0151 M

[Ag^{+}] = 0.0152*2 = 0.0304 M

Now, let's write the half reactions:

Cu ----------> Cu^{+} + 1e^{-} E° =
-0.52 V

Ag^{+} + 1e^{-} ---------> Ag E° = 0.80 V

E° = 0.80-0.52 = 0.28 V

Finally, calculate the cell potential with the nerst
equation:

E = E° - 0.059/n logK

E = 0.28 - 0.059log(1/0.0304)

E = 0.28 - 0.09

E = 0.19 V

Hope this helps

Given a potential of 0.456 V, for the following electrochemical
cell, what is the concentration of Ag+?
Cu2+(aq) (1.0 M)|Cu(s)||Ag+(aq)|Ag(s)

Predict the potential for the following voltaic cell at 25 C:
Cu(s)|CuSO4(0.42 M)||AgNO3(0.46 M)|Ag(s) (answer in volts)

A voltaic cell is constructed that uses the following reactions
and operates at 298K:
2Ag+(aq) + Cu(s) → 2Ag(s) +
Cu2+(aq)
If the cell generates an emf = +0.493 V and the [Ag+] = 0.795
M, what is the [Cu2+] ?

Calculate the cell potential for the voltaic cell based on the
following half reactions at T = 25ºC:
Cr3+(aq) + 3e- → Cr(s) Eº = - 0.74 V
TiO2+(aq) + 2H+ (aq) + 1e- → Ti3+(aq) + H2O(l) Eº = + 0.10 V
Where, [Cr3+] = 1.0 x 10-4 M, [TiO2+] = 1.0 x 10-1 M, [H+] = 1.0
M, [Ti3+] = 5.0 x 10-2 M.

The following galvanic cell has a potential of 0.578 v at 25
degress C: Ag(s) | AgCl(s) | Cl- (1.0 M) | |Ag+ (1.0 M) | Ag(s).
Use this information to calculate Ksp for AgCl at 25 degrees C.
Please answer the following questions:
The constant 0.0592 volts combines 2.303, __________
J/mole/K for R, __________ K for temperature, and __________C/mole
for F.
The value for the solubility product constant is
___________ (2 significant figures)

Please show all work and derivations!!
A voltaic cell is made by combining the following two
half-reactions:
Sn^(2+) (aq) + 2e- --> Sn(s) Eo red= +0.15 V
AgCl(s) + e- --> Ag(s) + Cl^- (aq) Eo red= +0.22 V
Calculate the cell potential E at 298K if [Sn^(2+)] = 0.050 M
and [Cl-] = 4.0 M.

A (2.1x10^-1) M Na2SO4 is also saturated with Ag2SO4. The [Ag+]
is (9.30x10^-3) M. What is the value of Ksp for Ag2SO4 at this
temperature?

1) This is voltaic cell Pt (s) | Pt2+ (aq, 5.0 M) || Au+ (aq,
2.0 × 10–5 M) | Au (s) How do I get cell potential of this voltaic
cell at 25°C?
These are the answer options : 0.360V, 0.640 V, 0.490V, 0.191V,
0.330V
2) What final concentration of NH3 is needed to lower Cu2+ to
6.4*10^-14M that was 1.43*10^4M in Cu^2+(aq)? (volume stays same
even NH3 gas is going through the solution) (Kf of [Cu(NH3)4] 2+ =...

A voltaic cell consists of a Pb/Pb+2 half-cell and a
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are [Pb+2]=0.0500 M and [Cu+2]=1.50 M. What
is the initial cell potential? What are the concentrations of
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Please show work
Calculate the cell potential for the following reaction that
takes place in an electrochemical cell at 25°C. Mn(s) ∣ Mn2+(aq,
1.28 M) ∥ Ag+(aq, 0.000837 M) ∣ Ag(s) Calculate the cell potential
for the following reaction that takes place in an electrochemical
cell at 25°C. Mn(s) Mn2+(aq, 1.28 M) Ag+(aq, 0.000837 M) Ag(s)
1.98 V
0.00 V
1.79 V
-0.84 V
-1.28 V

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