Question

For the following concentration cell: Mg(s) |
Mg^{2+}(0.24 M) | Mg^{2+}(0.53 M) | Mg(s)

(a) Determine the initial emf (Δε) for this cell (in volts)

(b) Determine the concentration of Mg^{2+} in each cell
when the emf of this cell goes to 0 volts

T = 298 K T = 310 K

Answer #1

For the following concentration cell: Mg(s) |
Mg^{2+}(0.24 M) | Mg^{2+}(0.53 M) | Mg(s)

(a) Determine the initial emf (Δε) for this cell (in volts)

(b) Determine the concentration of Mg^{2+} in each cell
when the emf of this cell goes to 0 volts

T = 298 K T = 310 K

We use the Nernst Equation:

E cell = E*cell - (RT/nF)(ln [dilute]/[concentrated])

E cell = 0 - [(8.314 J/mol-K)(298) / (2 mol e-)(96500 J/V-mol e-)]
x [ln (0.24/0.53)]

E cell = - (0.01284) (-0.79 )

=+ 0.0101 V

(b) Determine the concentration of Mg^{2+} in each cell
when the emf of this cell goes to 0 volts

T = 298 K T = 310 K

Then the concentration in [dilute] = [concentrated

At T = 298 K

We use the Nernst Equation:

E cell = E*cell - (RT/nF)(ln [dilute]/[concentrated])

E cell = 0 - [(8.314 J/mol-K)(298) / (2 mol e-)(96500 J/V-mol e-)]
x [ln 1.0]

E cell = 0 - [(8.314 J/mol-K)(298) / (2 mol e-)(96500 J/V-mol e-)] x 0

= 0

Calculate the emf of the following concentration cell: Mg(s) |
Mg2+(0.32 M) || Mg2+(0.45 M) | Mg(s) V

Enter your answer in the provided box.
Calculate the emf of the following concentration cell:
Mg(s) | Mg2+(0.33 M) ||
Mg2+(0.54 M) | Mg(s)
____ V

A concentration cell based on the following half reaction at 274
K
Mg2+ + 2 e- → Mg
SRP = -2.37 V
has initial concentrations of 1.31 M Mg2+, 0.239 M
Mg2+, and a potential of 0.02008 V at these conditions.
After 1.7 hours, the new potential of the cell is found to be
0.01032 V. What is the concentration of Mg2+ at the
cathode at this new potential?

What is the cell potential for the reaction
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 47 ∘C when [Fe2+]= 3.10 M and
[Mg2+]= 0.110 MM . Express the potential numerically in volts.

A concentration cell has both electrodes made of magnesium and
is based on the following half-cell reaction.
Mg2+(aq) + 2 e− → Mg(s)
(a) If [Mg2+ ] = 2.18 M at the cathode and [Mg2+ ] = 0.00744 M
at the anode, what is the cell potential of the concentration cell
at 298.15 K?
(b) What is the cell potential of the cell when the
concentration of Mg2+ ions is the same in the cathode
and anode compartments?

Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C ,
where [Fe2+]= 2.80 M and [Mg2+]= 0.210 M .
A. What is the value for the reaction quotient, Q, for the
cell?
B. What is the value for the temperature, T, in kelvins? What is
the value for n?
C. Calculate the cell potential.

A) When the Cu2+ concentration is
1.07 M, the observed cell potential at 298K for an
electrochemical cell with the following reaction is
1.607V. What is the
Mn2+ concentration?
Cu2+(aq) +
Mn(s)-->
Cu(s) +
Mn2+(aq)
Answer: _____ M
B) When the Cu2+ concentration is
5.71×10-4 M, the observed cell
potential at 298K for an electrochemical cell with the following
reaction is 2.609V. What is the
Mg2+ concentration?
Cu2+(aq) +
Mg(s)--->
Cu(s) +
Mg2+(aq)
Answer: _____ M
C) When the Cu2+...

What is the cell potential for the reaction
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 79 ∘C when [Fe2+]= 3.20 M and
[Mg2+]= 0.310 M .

What is the cell potential for the reaction
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 83 ∘C when [Fe2+]= 3.30 M and
[Mg2+]= 0.310 M .

When the Cu2+ concentration is
1.40 M, the observed cell potential at 298K for an
electrochemical cell with the following reaction is
2.805V. What is the
Mg2+ concentration?
Cu2+(aq) +
Mg(s) ----->
Cu(s) +
Mg2+(aq)
Answer: ______M

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